Calculus Examples

$lim_{t \to 0} \frac{9}{t \sqrt{1 + t}} - \frac{9}{t}$

Step 1

Combine terms.

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Step 1.1

To write $- \frac{9}{t}$ as a fraction with a common denominator, multiply by $\frac{\sqrt{1 + t}}{\sqrt{1 + t}}$ .

$lim_{t \to 0} \frac{9}{t \sqrt{1 + t}} - \frac{9}{t} \cdot \frac{\sqrt{1 + t}}{\sqrt{1 + t}}$

Step 1.2

Multiply $\frac{9}{t}$ by $\frac{\sqrt{1 + t}}{\sqrt{1 + t}}$ .

$lim_{t \to 0} \frac{9}{t \sqrt{1 + t}} - \frac{9 \sqrt{1 + t}}{t \sqrt{1 + t}}$

Step 1.3

Combine the numerators over the common denominator.

$lim_{t \to 0} \frac{9 - 9 \sqrt{1 + t}}{t \sqrt{1 + t}}$

Step 2

Apply L'Hospital's rule.

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Step 2.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 2.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{t \to 0} 9 - 9 \sqrt{1 + t}}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

Evaluate the limit.

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Step 2.1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $0$ .

$\frac{lim_{t \to 0} 9 - lim_{t \to 0} 9 \sqrt{1 + t}}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.2.1.2

Evaluate the limit of $9$ which is constant as $t$ approaches $0$ .

$\frac{9 - lim_{t \to 0} 9 \sqrt{1 + t}}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.2.1.3

Move the term $9$ outside of the limit because it is constant with respect to $t$ .

$\frac{9 - 9 lim_{t \to 0} \sqrt{1 + t}}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.2.1.4

Move the limit under the radical sign.

$\frac{9 - 9 \sqrt{lim_{t \to 0} 1 + t}}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.2.1.5

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $0$ .

$\frac{9 - 9 \sqrt{lim_{t \to 0} 1 + lim_{t \to 0} t}}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.2.1.6

Evaluate the limit of $1$ which is constant as $t$ approaches $0$ .

$\frac{9 - 9 \sqrt{1 + lim_{t \to 0} t}}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.2.2

Evaluate the limit of $t$ by plugging in $0$ for $t$ .

$\frac{9 - 9 \sqrt{1 + 0}}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.2.3

Simplify the answer.

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Step 2.1.2.3.1

Simplify each term.

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Step 2.1.2.3.1.1

Add $1$ and $0$ .

$\frac{9 - 9 \sqrt{1}}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.2.3.1.2

Any root of $1$ is $1$ .

$\frac{9 - 9 \cdot 1}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.2.3.1.3

Multiply $- 9$ by $1$ .

$\frac{9 - 9}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.2.3.2

Subtract $9$ from $9$ .

$\frac{0}{lim_{t \to 0} t \sqrt{1 + t}}$

Step 2.1.3

Evaluate the limit of the denominator.

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Step 2.1.3.1

Split the limit using the Product of Limits Rule on the limit as $t$ approaches $0$ .

$\frac{0}{lim_{t \to 0} t \cdot lim_{t \to 0} \sqrt{1 + t}}$

Step 2.1.3.2

Move the limit under the radical sign.

$\frac{0}{lim_{t \to 0} t \cdot \sqrt{lim_{t \to 0} 1 + t}}$

Step 2.1.3.3

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $0$ .

$\frac{0}{lim_{t \to 0} t \cdot \sqrt{lim_{t \to 0} 1 + lim_{t \to 0} t}}$

Step 2.1.3.4

Evaluate the limit of $1$ which is constant as $t$ approaches $0$ .

$\frac{0}{lim_{t \to 0} t \cdot \sqrt{1 + lim_{t \to 0} t}}$

Step 2.1.3.5

Evaluate the limits by plugging in $0$ for all occurrences of $t$ .

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Step 2.1.3.5.1

Evaluate the limit of $t$ by plugging in $0$ for $t$ .

$\frac{0}{0 \cdot \sqrt{1 + lim_{t \to 0} t}}$

Step 2.1.3.5.2

Evaluate the limit of $t$ by plugging in $0$ for $t$ .

$\frac{0}{0 \cdot \sqrt{1 + 0}}$

Step 2.1.3.6

Simplify the answer.

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Step 2.1.3.6.1

Add $1$ and $0$ .

$\frac{0}{0 \cdot \sqrt{1}}$

Step 2.1.3.6.2

Any root of $1$ is $1$ .

$\frac{0}{0 \cdot 1}$

Step 2.1.3.6.3

Multiply $0$ by $1$ .

$\frac{0}{0}$

Step 2.1.3.6.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 2.1.3.7

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 2.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 2.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{t \to 0} \frac{9 - 9 \sqrt{1 + t}}{t \sqrt{1 + t}} = lim_{t \to 0} \frac{\frac{d}{d t} [9 - 9 \sqrt{1 + t}]}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$lim_{t \to 0} \frac{\frac{d}{d t} [9 - 9 \sqrt{1 + t}]}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.2

By the Sum Rule, the derivative of $9 - 9 \sqrt{1 + t}$ with respect to $t$ is $\frac{d}{d t} [9] + \frac{d}{d t} [- 9 \sqrt{1 + t}]$ .

$lim_{t \to 0} \frac{\frac{d}{d t} [9] + \frac{d}{d t} [- 9 \sqrt{1 + t}]}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.3

Since $9$ is constant with respect to $t$ , the derivative of $9$ with respect to $t$ is $0$ .

$lim_{t \to 0} \frac{0 + \frac{d}{d t} [- 9 \sqrt{1 + t}]}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4

Evaluate $\frac{d}{d t} [- 9 \sqrt{1 + t}]$ .

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Step 2.3.4.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{1 + t}$ as ${(1 + t)}^{\frac{1}{2}}$ .

$lim_{t \to 0} \frac{0 + \frac{d}{d t} [- 9 {(1 + t)}^{\frac{1}{2}}]}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.2

Since $- 9$ is constant with respect to $t$ , the derivative of $- 9 {(1 + t)}^{\frac{1}{2}}$ with respect to $t$ is $- 9 \frac{d}{d t} [{(1 + t)}^{\frac{1}{2}}]$ .

$lim_{t \to 0} \frac{0 - 9 \frac{d}{d t} [{(1 + t)}^{\frac{1}{2}}]}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{\frac{1}{2}}$ and $g (t) = 1 + t$ .

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Step 2.3.4.3.1

To apply the Chain Rule, set $u$ as $1 + t$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d t} [1 + t])}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{2}$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d t} [1 + t])}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.3.3

Replace all occurrences of $u$ with $1 + t$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} {(1 + t)}^{\frac{1}{2} - 1} \frac{d}{d t} [1 + t])}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.4

By the Sum Rule, the derivative of $1 + t$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [t]$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} {(1 + t)}^{\frac{1}{2} - 1} (\frac{d}{d t} [1] + \frac{d}{d t} [t]))}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.5

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} {(1 + t)}^{\frac{1}{2} - 1} (0 + \frac{d}{d t} [t]))}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.6

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} {(1 + t)}^{\frac{1}{2} - 1} (0 + 1))}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.7

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} {(1 + t)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (0 + 1))}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.8

Combine $- 1$ and $\frac{2}{2}$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} {(1 + t)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (0 + 1))}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.9

Combine the numerators over the common denominator.

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} {(1 + t)}^{\frac{1 - 1 \cdot 2}{2}} (0 + 1))}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.10

Simplify the numerator.

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Step 2.3.4.10.1

Multiply $- 1$ by $2$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} {(1 + t)}^{\frac{1 - 2}{2}} (0 + 1))}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.10.2

Subtract $2$ from $1$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} {(1 + t)}^{\frac{- 1}{2}} (0 + 1))}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.11

Move the negative in front of the fraction.

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} {(1 + t)}^{- \frac{1}{2}} (0 + 1))}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.12

Add $0$ and $1$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{1}{2} {(1 + t)}^{- \frac{1}{2}} \cdot 1)}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.13

Combine $\frac{1}{2}$ and ${(1 + t)}^{- \frac{1}{2}}$ .

$lim_{t \to 0} \frac{0 - 9 (\frac{{(1 + t)}^{- \frac{1}{2}}}{2} \cdot 1)}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.14

Multiply $\frac{{(1 + t)}^{- \frac{1}{2}}}{2}$ by $1$ .

$lim_{t \to 0} \frac{0 - 9 \frac{{(1 + t)}^{- \frac{1}{2}}}{2}}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.15

Move ${(1 + t)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{t \to 0} \frac{0 - 9 \frac{1}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.16

Combine $- 9$ and $\frac{1}{2 {(1 + t)}^{\frac{1}{2}}}$ .

$lim_{t \to 0} \frac{0 + \frac{- 9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.4.17

Move the negative in front of the fraction.

$lim_{t \to 0} \frac{0 - \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.5

Subtract $\frac{9}{2 {(1 + t)}^{\frac{1}{2}}}$ from $0$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{d}{d t} [t \sqrt{1 + t}]}$

Step 2.3.6

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{1 + t}$ as ${(1 + t)}^{\frac{1}{2}}$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{d}{d t} [t {(1 + t)}^{\frac{1}{2}}]}$

Step 2.3.7

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t$ and $g (t) = {(1 + t)}^{\frac{1}{2}}$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t \frac{d}{d t} [{(1 + t)}^{\frac{1}{2}}] + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.8

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{\frac{1}{2}}$ and $g (t) = 1 + t$ .

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Step 2.3.8.1

To apply the Chain Rule, set $u$ as $1 + t$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t (\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d t} [1 + t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.8.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{2}$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d t} [1 + t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.8.3

Replace all occurrences of $u$ with $1 + t$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t (\frac{1}{2} {(1 + t)}^{\frac{1}{2} - 1} \frac{d}{d t} [1 + t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.9

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t (\frac{1}{2} {(1 + t)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d t} [1 + t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.10

Combine $- 1$ and $\frac{2}{2}$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t (\frac{1}{2} {(1 + t)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d t} [1 + t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.11

Combine the numerators over the common denominator.

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t (\frac{1}{2} {(1 + t)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d t} [1 + t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.12

Simplify the numerator.

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Step 2.3.12.1

Multiply $- 1$ by $2$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t (\frac{1}{2} {(1 + t)}^{\frac{1 - 2}{2}} \frac{d}{d t} [1 + t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.12.2

Subtract $2$ from $1$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t (\frac{1}{2} {(1 + t)}^{\frac{- 1}{2}} \frac{d}{d t} [1 + t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.13

Move the negative in front of the fraction.

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t (\frac{1}{2} {(1 + t)}^{- \frac{1}{2}} \frac{d}{d t} [1 + t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.14

Combine $\frac{1}{2}$ and ${(1 + t)}^{- \frac{1}{2}}$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t (\frac{{(1 + t)}^{- \frac{1}{2}}}{2} \frac{d}{d t} [1 + t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.15

Move ${(1 + t)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{t (\frac{1}{2 {(1 + t)}^{\frac{1}{2}}} \frac{d}{d t} [1 + t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.16

Combine $\frac{1}{2 {(1 + t)}^{\frac{1}{2}}}$ and $t$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t}{2 {(1 + t)}^{\frac{1}{2}}} \frac{d}{d t} [1 + t] + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.17

By the Sum Rule, the derivative of $1 + t$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [t]$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t}{2 {(1 + t)}^{\frac{1}{2}}} (\frac{d}{d t} [1] + \frac{d}{d t} [t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.18

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t}{2 {(1 + t)}^{\frac{1}{2}}} (0 + \frac{d}{d t} [t]) + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.19

Add $0$ and $\frac{d}{d t} [t]$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t}{2 {(1 + t)}^{\frac{1}{2}}} \frac{d}{d t} [t] + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.20

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t}{2 {(1 + t)}^{\frac{1}{2}}} \cdot 1 + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.21

Multiply $\frac{t}{2 {(1 + t)}^{\frac{1}{2}}}$ by $1$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t}{2 {(1 + t)}^{\frac{1}{2}}} + {(1 + t)}^{\frac{1}{2}} \frac{d}{d t} [t]}$

Step 2.3.22

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t}{2 {(1 + t)}^{\frac{1}{2}}} + {(1 + t)}^{\frac{1}{2}} \cdot 1}$

Step 2.3.23

Multiply ${(1 + t)}^{\frac{1}{2}}$ by $1$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t}{2 {(1 + t)}^{\frac{1}{2}}} + {(1 + t)}^{\frac{1}{2}}}$

Step 2.3.24

To write ${(1 + t)}^{\frac{1}{2}}$ as a fraction with a common denominator, multiply by $\frac{2 {(1 + t)}^{\frac{1}{2}}}{2 {(1 + t)}^{\frac{1}{2}}}$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t}{2 {(1 + t)}^{\frac{1}{2}}} + {(1 + t)}^{\frac{1}{2}} \cdot \frac{2 {(1 + t)}^{\frac{1}{2}}}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.25

Combine ${(1 + t)}^{\frac{1}{2}}$ and $\frac{2 {(1 + t)}^{\frac{1}{2}}}{2 {(1 + t)}^{\frac{1}{2}}}$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t}{2 {(1 + t)}^{\frac{1}{2}}} + \frac{{(1 + t)}^{\frac{1}{2}} (2 {(1 + t)}^{\frac{1}{2}})}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.26

Combine the numerators over the common denominator.

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t + {(1 + t)}^{\frac{1}{2}} (2 {(1 + t)}^{\frac{1}{2}})}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.27

Multiply ${(1 + t)}^{\frac{1}{2}}$ by ${(1 + t)}^{\frac{1}{2}}$ by adding the exponents.

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Step 2.3.27.1

Move ${(1 + t)}^{\frac{1}{2}}$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t + {(1 + t)}^{\frac{1}{2}} {(1 + t)}^{\frac{1}{2}} \cdot 2}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.27.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t + {(1 + t)}^{\frac{1}{2} + \frac{1}{2}} \cdot 2}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.27.3

Combine the numerators over the common denominator.

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t + {(1 + t)}^{\frac{1 + 1}{2}} \cdot 2}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.27.4

Add $1$ and $1$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t + {(1 + t)}^{\frac{2}{2}} \cdot 2}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.27.5

Divide $2$ by $2$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t + {(1 + t)}^{1} \cdot 2}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.28

Simplify ${(1 + t)}^{1} \cdot 2$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t + (1 + t) \cdot 2}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.29

Move $2$ to the left of $1 + t$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t + 2 \cdot (1 + t)}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.30

Simplify.

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Step 2.3.30.1

Apply the distributive property.

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t + 2 \cdot 1 + 2 t}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.30.2

Simplify the numerator.

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Step 2.3.30.2.1

Multiply $2$ by $1$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{t + 2 + 2 t}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.3.30.2.2

Add $t$ and $2 t$ .

$lim_{t \to 0} \frac{- \frac{9}{2 {(1 + t)}^{\frac{1}{2}}}}{\frac{3 t + 2}{2 {(1 + t)}^{\frac{1}{2}}}}$

Step 2.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{t \to 0} - \frac{9}{2 {(1 + t)}^{\frac{1}{2}}} \cdot \frac{2 {(1 + t)}^{\frac{1}{2}}}{3 t + 2}$

Step 2.5

Convert fractional exponents to radicals.

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Step 2.5.1

Rewrite ${(1 + t)}^{\frac{1}{2}}$ as $\sqrt{1 + t}$ .

$lim_{t \to 0} - \frac{9}{2 \sqrt{1 + t}} \cdot \frac{2 {(1 + t)}^{\frac{1}{2}}}{3 t + 2}$

Step 2.5.2

Rewrite ${(1 + t)}^{\frac{1}{2}}$ as $\sqrt{1 + t}$ .

$lim_{t \to 0} - \frac{9}{2 \sqrt{1 + t}} \cdot \frac{2 \sqrt{1 + t}}{3 t + 2}$

Step 2.6

Combine factors.

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Step 2.6.1

Multiply $\frac{2 \sqrt{1 + t}}{3 t + 2}$ by $\frac{9}{2 \sqrt{1 + t}}$ .

$lim_{t \to 0} - \frac{2 \sqrt{1 + t} \cdot 9}{(3 t + 2) (2 \sqrt{1 + t})}$

Step 2.6.2

Multiply $9$ by $2$ .

$lim_{t \to 0} - \frac{18 \sqrt{1 + t}}{(3 t + 2) (2 \sqrt{1 + t})}$

Step 2.7

Reduce.

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Step 2.7.1

Cancel the common factor of $18$ and $2$ .

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Step 2.7.1.1

Factor $2$ out of $18 \sqrt{1 + t}$ .

$lim_{t \to 0} - \frac{2 (9 \sqrt{1 + t})}{(3 t + 2) (2 \sqrt{1 + t})}$

Step 2.7.1.2

Cancel the common factors.

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Step 2.7.1.2.1

Factor $2$ out of $(3 t + 2) (2 \sqrt{1 + t})$ .

$lim_{t \to 0} - \frac{2 (9 \sqrt{1 + t})}{2 ((3 t + 2) (\sqrt{1 + t}))}$

Step 2.7.1.2.2

Cancel the common factor.

$lim_{t \to 0} - \frac{2 (9 \sqrt{1 + t})}{2 ((3 t + 2) (\sqrt{1 + t}))}$

Step 2.7.1.2.3

Rewrite the expression.

$lim_{t \to 0} - \frac{9 \sqrt{1 + t}}{(3 t + 2) (\sqrt{1 + t})}$

Step 2.7.2

Cancel the common factor of $\sqrt{1 + t}$ .

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Step 2.7.2.1

Cancel the common factor.

$lim_{t \to 0} - \frac{9 \sqrt{1 + t}}{(3 t + 2) \sqrt{1 + t}}$

Step 2.7.2.2

Rewrite the expression.

$lim_{t \to 0} - \frac{9}{3 t + 2}$

Step 3

Evaluate the limit.

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Step 3.1

Move the term $- 1$ outside of the limit because it is constant with respect to $t$ .

$- lim_{t \to 0} \frac{9}{3 t + 2}$

Step 3.2

Move the term $9$ outside of the limit because it is constant with respect to $t$ .

$- 9 lim_{t \to 0} \frac{1}{3 t + 2}$

Step 3.3

Split the limit using the Limits Quotient Rule on the limit as $t$ approaches $0$ .

$- 9 \frac{lim_{t \to 0} 1}{lim_{t \to 0} 3 t + 2}$

Step 3.4

Evaluate the limit of $1$ which is constant as $t$ approaches $0$ .

$- 9 \frac{1}{lim_{t \to 0} 3 t + 2}$

Step 3.5

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $0$ .

$- 9 \frac{1}{lim_{t \to 0} 3 t + lim_{t \to 0} 2}$

Step 3.6

Move the term $3$ outside of the limit because it is constant with respect to $t$ .

$- 9 \frac{1}{3 lim_{t \to 0} t + lim_{t \to 0} 2}$

Step 3.7

Evaluate the limit of $2$ which is constant as $t$ approaches $0$ .

$- 9 \frac{1}{3 lim_{t \to 0} t + 2}$

Step 4

Evaluate the limit of $t$ by plugging in $0$ for $t$ .

$- 9 \frac{1}{3 \cdot 0 + 2}$

Step 5

Simplify the answer.

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Step 5.1

Simplify the denominator.

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Step 5.1.1

Multiply $3$ by $0$ .

$- 9 \frac{1}{0 + 2}$

Step 5.1.2

Add $0$ and $2$ .

$- 9 (\frac{1}{2})$

Step 5.2

Combine $- 9$ and $\frac{1}{2}$ .

$\frac{- 9}{2}$

Step 5.3

Move the negative in front of the fraction.

$- \frac{9}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$- \frac{9}{2}$

Decimal Form:

$- 4.5$