Calculus Examples

$lim_{x \to 2} \frac{x^{2} + 3 x - 10}{\sqrt{4 x - 4} - x}$

Step 1

Apply L'Hospital's rule.

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Step 1.1

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

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Step 1.1.1

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

$\frac{lim_{x \to 2} x^{2} + 3 x - 10}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{lim_{x \to 2} x^{2} + lim_{x \to 2} 3 x - lim_{x \to 2} 10}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.2.2

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

$\frac{{(lim_{x \to 2} x)}^{2} + lim_{x \to 2} 3 x - lim_{x \to 2} 10}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.2.3

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

$\frac{{(lim_{x \to 2} x)}^{2} + 3 lim_{x \to 2} x - lim_{x \to 2} 10}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.2.4

Evaluate the limit of $10$ which is constant as $x$ approaches $2$ .

$\frac{{(lim_{x \to 2} x)}^{2} + 3 lim_{x \to 2} x - 1 \cdot 10}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.2.5

Evaluate the limits by plugging in $2$ for all occurrences of $x$ .

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Step 1.1.2.5.1

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{2^{2} + 3 lim_{x \to 2} x - 1 \cdot 10}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.2.5.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{2^{2} + 3 \cdot 2 - 1 \cdot 10}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.2.6

Simplify the answer.

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Step 1.1.2.6.1

Simplify each term.

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Step 1.1.2.6.1.1

Raise $2$ to the power of $2$ .

$\frac{4 + 3 \cdot 2 - 1 \cdot 10}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.2.6.1.2

Multiply $3$ by $2$ .

$\frac{4 + 6 - 1 \cdot 10}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.2.6.1.3

Multiply $- 1$ by $10$ .

$\frac{4 + 6 - 10}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.2.6.2

Add $4$ and $6$ .

$\frac{10 - 10}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.2.6.3

Subtract $10$ from $10$ .

$\frac{0}{lim_{x \to 2} \sqrt{4 x - 4} - x}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{0}{lim_{x \to 2} \sqrt{4 x - 4} - lim_{x \to 2} x}$

Step 1.1.3.2

Move the limit under the radical sign.

$\frac{0}{\sqrt{lim_{x \to 2} 4 x - 4} - lim_{x \to 2} x}$

Step 1.1.3.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{0}{\sqrt{lim_{x \to 2} 4 x - lim_{x \to 2} 4} - lim_{x \to 2} x}$

Step 1.1.3.4

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{\sqrt{4 lim_{x \to 2} x - lim_{x \to 2} 4} - lim_{x \to 2} x}$

Step 1.1.3.5

Evaluate the limit of $4$ which is constant as $x$ approaches $2$ .

$\frac{0}{\sqrt{4 lim_{x \to 2} x - 1 \cdot 4} - lim_{x \to 2} x}$

Step 1.1.3.6

Evaluate the limits by plugging in $2$ for all occurrences of $x$ .

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Step 1.1.3.6.1

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{0}{\sqrt{4 \cdot 2 - 1 \cdot 4} - lim_{x \to 2} x}$

Step 1.1.3.6.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{0}{\sqrt{4 \cdot 2 - 1 \cdot 4} - 2}$

Step 1.1.3.7

Simplify the answer.

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Step 1.1.3.7.1

Simplify each term.

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Step 1.1.3.7.1.1

Multiply $4$ by $2$ .

$\frac{0}{\sqrt{8 - 1 \cdot 4} - 2}$

Step 1.1.3.7.1.2

Multiply $- 1$ by $4$ .

$\frac{0}{\sqrt{8 - 4} - 2}$

Step 1.1.3.7.1.3

Subtract $4$ from $8$ .

$\frac{0}{\sqrt{4} - 2}$

Step 1.1.3.7.1.4

Rewrite $4$ as $2^{2}$ .

$\frac{0}{\sqrt{2^{2}} - 2}$

Step 1.1.3.7.1.5

Pull terms out from under the radical, assuming positive real numbers.

$\frac{0}{2 - 2}$

Step 1.1.3.7.2

Subtract $2$ from $2$ .

$\frac{0}{0}$

Step 1.1.3.7.3

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

Undefined

$\frac{0}{0}$

Step 1.1.3.8

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

Undefined

$\frac{0}{0}$

Step 1.1.4

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

Undefined

$\frac{0}{0}$

Step 1.2

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.

$lim_{x \to 2} \frac{x^{2} + 3 x - 10}{\sqrt{4 x - 4} - x} = lim_{x \to 2} \frac{\frac{d}{d x} [x^{2} + 3 x - 10]}{\frac{d}{d x} [\sqrt{4 x - 4} - x]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 2} \frac{\frac{d}{d x} [x^{2} + 3 x - 10]}{\frac{d}{d x} [\sqrt{4 x - 4} - x]}$

Step 1.3.2

By the Sum Rule, the derivative of $x^{2} + 3 x - 10$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [- 10]$ .

$lim_{x \to 2} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [- 10]}{\frac{d}{d x} [\sqrt{4 x - 4} - x]}$

Step 1.3.3

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

$lim_{x \to 2} \frac{2 x + \frac{d}{d x} [3 x] + \frac{d}{d x} [- 10]}{\frac{d}{d x} [\sqrt{4 x - 4} - x]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [3 x]$ .

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Step 1.3.4.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$lim_{x \to 2} \frac{2 x + 3 \frac{d}{d x} [x] + \frac{d}{d x} [- 10]}{\frac{d}{d x} [\sqrt{4 x - 4} - x]}$

Step 1.3.4.2

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

$lim_{x \to 2} \frac{2 x + 3 \cdot 1 + \frac{d}{d x} [- 10]}{\frac{d}{d x} [\sqrt{4 x - 4} - x]}$

Step 1.3.4.3

Multiply $3$ by $1$ .

$lim_{x \to 2} \frac{2 x + 3 + \frac{d}{d x} [- 10]}{\frac{d}{d x} [\sqrt{4 x - 4} - x]}$

Step 1.3.5

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10$ with respect to $x$ is $0$ .

$lim_{x \to 2} \frac{2 x + 3 + 0}{\frac{d}{d x} [\sqrt{4 x - 4} - x]}$

Step 1.3.6

Add $2 x + 3$ and $0$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{d}{d x} [\sqrt{4 x - 4} - x]}$

Step 1.3.7

By the Sum Rule, the derivative of $\sqrt{4 x - 4} - x$ with respect to $x$ is $\frac{d}{d x} [\sqrt{4 x - 4}] + \frac{d}{d x} [- x]$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{d}{d x} [\sqrt{4 x - 4}] + \frac{d}{d x} [- x]}$

Step 1.3.8

Evaluate $\frac{d}{d x} [\sqrt{4 x - 4}]$ .

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Step 1.3.8.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 x - 4}$ as ${(4 x - 4)}^{\frac{1}{2}}$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{d}{d x} [{(4 x - 4)}^{\frac{1}{2}}] + \frac{d}{d x} [- x]}$

Step 1.3.8.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = 4 x - 4$ .

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Step 1.3.8.2.1

To apply the Chain Rule, set $u$ as $4 x - 4$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [4 x - 4] + \frac{d}{d x} [- x]}$

Step 1.3.8.2.2

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

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [4 x - 4] + \frac{d}{d x} [- x]}$

Step 1.3.8.2.3

Replace all occurrences of $u$ with $4 x - 4$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{\frac{1}{2} - 1} \frac{d}{d x} [4 x - 4] + \frac{d}{d x} [- x]}$

Step 1.3.8.3

By the Sum Rule, the derivative of $4 x - 4$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [- 4]$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{\frac{1}{2} - 1} (\frac{d}{d x} [4 x] + \frac{d}{d x} [- 4]) + \frac{d}{d x} [- x]}$

Step 1.3.8.4

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{\frac{1}{2} - 1} (4 \frac{d}{d x} [x] + \frac{d}{d x} [- 4]) + \frac{d}{d x} [- x]}$

Step 1.3.8.5

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

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{\frac{1}{2} - 1} (4 \cdot 1 + \frac{d}{d x} [- 4]) + \frac{d}{d x} [- x]}$

Step 1.3.8.6

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{\frac{1}{2} - 1} (4 \cdot 1 + 0) + \frac{d}{d x} [- x]}$

Step 1.3.8.7

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

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (4 \cdot 1 + 0) + \frac{d}{d x} [- x]}$

Step 1.3.8.8

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

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (4 \cdot 1 + 0) + \frac{d}{d x} [- x]}$

Step 1.3.8.9

Combine the numerators over the common denominator.

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{\frac{1 - 1 \cdot 2}{2}} (4 \cdot 1 + 0) + \frac{d}{d x} [- x]}$

Step 1.3.8.10

Simplify the numerator.

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Step 1.3.8.10.1

Multiply $- 1$ by $2$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{\frac{1 - 2}{2}} (4 \cdot 1 + 0) + \frac{d}{d x} [- x]}$

Step 1.3.8.10.2

Subtract $2$ from $1$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{\frac{- 1}{2}} (4 \cdot 1 + 0) + \frac{d}{d x} [- x]}$

Step 1.3.8.11

Move the negative in front of the fraction.

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{- \frac{1}{2}} (4 \cdot 1 + 0) + \frac{d}{d x} [- x]}$

Step 1.3.8.12

Multiply $4$ by $1$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{- \frac{1}{2}} (4 + 0) + \frac{d}{d x} [- x]}$

Step 1.3.8.13

Add $4$ and $0$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{1}{2} {(4 x - 4)}^{- \frac{1}{2}} \cdot 4 + \frac{d}{d x} [- x]}$

Step 1.3.8.14

Combine $\frac{1}{2}$ and ${(4 x - 4)}^{- \frac{1}{2}}$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{{(4 x - 4)}^{- \frac{1}{2}}}{2} \cdot 4 + \frac{d}{d x} [- x]}$

Step 1.3.8.15

Combine $\frac{{(4 x - 4)}^{- \frac{1}{2}}}{2}$ and $4$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{{(4 x - 4)}^{- \frac{1}{2}} \cdot 4}{2} + \frac{d}{d x} [- x]}$

Step 1.3.8.16

Move $4$ to the left of ${(4 x - 4)}^{- \frac{1}{2}}$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{4 \cdot {(4 x - 4)}^{- \frac{1}{2}}}{2} + \frac{d}{d x} [- x]}$

Step 1.3.8.17

Move ${(4 x - 4)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{4}{2 {(4 x - 4)}^{\frac{1}{2}}} + \frac{d}{d x} [- x]}$

Step 1.3.8.18

Factor $2$ out of $4$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{2 \cdot 2}{2 {(4 x - 4)}^{\frac{1}{2}}} + \frac{d}{d x} [- x]}$

Step 1.3.8.19

Cancel the common factors.

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Step 1.3.8.19.1

Factor $2$ out of $2 {(4 x - 4)}^{\frac{1}{2}}$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{2 \cdot 2}{2 ({(4 x - 4)}^{\frac{1}{2}})} + \frac{d}{d x} [- x]}$

Step 1.3.8.19.2

Cancel the common factor.

$lim_{x \to 2} \frac{2 x + 3}{\frac{2 \cdot 2}{2 {(4 x - 4)}^{\frac{1}{2}}} + \frac{d}{d x} [- x]}$

Step 1.3.8.19.3

Rewrite the expression.

$lim_{x \to 2} \frac{2 x + 3}{\frac{2}{{(4 x - 4)}^{\frac{1}{2}}} + \frac{d}{d x} [- x]}$

Step 1.3.9

Evaluate $\frac{d}{d x} [- x]$ .

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Step 1.3.9.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{2}{{(4 x - 4)}^{\frac{1}{2}}} - \frac{d}{d x} [x]}$

Step 1.3.9.2

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

$lim_{x \to 2} \frac{2 x + 3}{\frac{2}{{(4 x - 4)}^{\frac{1}{2}}} - 1 \cdot 1}$

Step 1.3.9.3

Multiply $- 1$ by $1$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{2}{{(4 x - 4)}^{\frac{1}{2}}} - 1}$

Step 1.4

Rewrite ${(4 x - 4)}^{\frac{1}{2}}$ as $\sqrt{4 x - 4}$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{2}{\sqrt{4 x - 4}} - 1}$

Step 1.5

Combine terms.

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Step 1.5.1

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{\sqrt{4 x - 4}}{\sqrt{4 x - 4}}$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{2}{\sqrt{4 x - 4}} - 1 \cdot \frac{\sqrt{4 x - 4}}{\sqrt{4 x - 4}}}$

Step 1.5.2

Combine $- 1$ and $\frac{\sqrt{4 x - 4}}{\sqrt{4 x - 4}}$ .

$lim_{x \to 2} \frac{2 x + 3}{\frac{2}{\sqrt{4 x - 4}} + \frac{- \sqrt{4 x - 4}}{\sqrt{4 x - 4}}}$

Step 1.5.3

Combine the numerators over the common denominator.

$lim_{x \to 2} \frac{2 x + 3}{\frac{2 - \sqrt{4 x - 4}}{\sqrt{4 x - 4}}}$

Step 2

Since the function approaches $\infty$ from the left and $- \infty$ from the right, the limit does not exist.

$Does not exist$