Calculus Examples

$lim_{t \to - 6} \frac{t^{2} - 36}{2 t^{2} + 13 t + 6}$

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.

$\frac{lim_{t \to - 6} t^{2} - 36}{lim_{t \to - 6} 2 t^{2} + 13 t + 6}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $- 6$ .

$\frac{lim_{t \to - 6} t^{2} - lim_{t \to - 6} 36}{lim_{t \to - 6} 2 t^{2} + 13 t + 6}$

Step 1.1.2.1.2

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

$\frac{{(lim_{t \to - 6} t)}^{2} - lim_{t \to - 6} 36}{lim_{t \to - 6} 2 t^{2} + 13 t + 6}$

Step 1.1.2.1.3

Evaluate the limit of $36$ which is constant as $t$ approaches $- 6$ .

$\frac{{(lim_{t \to - 6} t)}^{2} - 1 \cdot 36}{lim_{t \to - 6} 2 t^{2} + 13 t + 6}$

Step 1.1.2.2

Evaluate the limit of $t$ by plugging in $- 6$ for $t$ .

$\frac{{(- 6)}^{2} - 1 \cdot 36}{lim_{t \to - 6} 2 t^{2} + 13 t + 6}$

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

Raise $- 6$ to the power of $2$ .

$\frac{36 - 1 \cdot 36}{lim_{t \to - 6} 2 t^{2} + 13 t + 6}$

Step 1.1.2.3.1.2

Multiply $- 1$ by $36$ .

$\frac{36 - 36}{lim_{t \to - 6} 2 t^{2} + 13 t + 6}$

Step 1.1.2.3.2

Subtract $36$ from $36$ .

$\frac{0}{lim_{t \to - 6} 2 t^{2} + 13 t + 6}$

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 $t$ approaches $- 6$ .

$\frac{0}{lim_{t \to - 6} 2 t^{2} + lim_{t \to - 6} 13 t + lim_{t \to - 6} 6}$

Step 1.1.3.2

Move the term $2$ outside of the limit because it is constant with respect to $t$ .

$\frac{0}{2 lim_{t \to - 6} t^{2} + lim_{t \to - 6} 13 t + lim_{t \to - 6} 6}$

Step 1.1.3.3

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

$\frac{0}{2 {(lim_{t \to - 6} t)}^{2} + lim_{t \to - 6} 13 t + lim_{t \to - 6} 6}$

Step 1.1.3.4

Move the term $13$ outside of the limit because it is constant with respect to $t$ .

$\frac{0}{2 {(lim_{t \to - 6} t)}^{2} + 13 lim_{t \to - 6} t + lim_{t \to - 6} 6}$

Step 1.1.3.5

Evaluate the limit of $6$ which is constant as $t$ approaches $- 6$ .

$\frac{0}{2 {(lim_{t \to - 6} t)}^{2} + 13 lim_{t \to - 6} t + 6}$

Step 1.1.3.6

Evaluate the limits by plugging in $- 6$ for all occurrences of $t$ .

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Step 1.1.3.6.1

Evaluate the limit of $t$ by plugging in $- 6$ for $t$ .

$\frac{0}{2 {(- 6)}^{2} + 13 lim_{t \to - 6} t + 6}$

Step 1.1.3.6.2

Evaluate the limit of $t$ by plugging in $- 6$ for $t$ .

$\frac{0}{2 {(- 6)}^{2} + 13 \cdot - 6 + 6}$

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

Raise $- 6$ to the power of $2$ .

$\frac{0}{2 \cdot 36 + 13 \cdot - 6 + 6}$

Step 1.1.3.7.1.2

Multiply $2$ by $36$ .

$\frac{0}{72 + 13 \cdot - 6 + 6}$

Step 1.1.3.7.1.3

Multiply $13$ by $- 6$ .

$\frac{0}{72 - 78 + 6}$

Step 1.1.3.7.2

Subtract $78$ from $72$ .

$\frac{0}{- 6 + 6}$

Step 1.1.3.7.3

Add $- 6$ and $6$ .

$\frac{0}{0}$

Step 1.1.3.7.4

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.

$lim_{t \to - 6} \frac{t^{2} - 36}{2 t^{2} + 13 t + 6} = lim_{t \to - 6} \frac{\frac{d}{d t} [t^{2} - 36]}{\frac{d}{d t} [2 t^{2} + 13 t + 6]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{t \to - 6} \frac{\frac{d}{d t} [t^{2} - 36]}{\frac{d}{d t} [2 t^{2} + 13 t + 6]}$

Step 1.3.2

By the Sum Rule, the derivative of $t^{2} - 36$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [- 36]$ .

$lim_{t \to - 6} \frac{\frac{d}{d t} [t^{2}] + \frac{d}{d t} [- 36]}{\frac{d}{d t} [2 t^{2} + 13 t + 6]}$

Step 1.3.3

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

$lim_{t \to - 6} \frac{2 t + \frac{d}{d t} [- 36]}{\frac{d}{d t} [2 t^{2} + 13 t + 6]}$

Step 1.3.4

Since $- 36$ is constant with respect to $t$ , the derivative of $- 36$ with respect to $t$ is $0$ .

$lim_{t \to - 6} \frac{2 t + 0}{\frac{d}{d t} [2 t^{2} + 13 t + 6]}$

Step 1.3.5

Add $2 t$ and $0$ .

$lim_{t \to - 6} \frac{2 t}{\frac{d}{d t} [2 t^{2} + 13 t + 6]}$

Step 1.3.6

By the Sum Rule, the derivative of $2 t^{2} + 13 t + 6$ with respect to $t$ is $\frac{d}{d t} [2 t^{2}] + \frac{d}{d t} [13 t] + \frac{d}{d t} [6]$ .

$lim_{t \to - 6} \frac{2 t}{\frac{d}{d t} [2 t^{2}] + \frac{d}{d t} [13 t] + \frac{d}{d t} [6]}$

Step 1.3.7

Evaluate $\frac{d}{d t} [2 t^{2}]$ .

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Step 1.3.7.1

Since $2$ is constant with respect to $t$ , the derivative of $2 t^{2}$ with respect to $t$ is $2 \frac{d}{d t} [t^{2}]$ .

$lim_{t \to - 6} \frac{2 t}{2 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [13 t] + \frac{d}{d t} [6]}$

Step 1.3.7.2

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

$lim_{t \to - 6} \frac{2 t}{2 (2 t) + \frac{d}{d t} [13 t] + \frac{d}{d t} [6]}$

Step 1.3.7.3

Multiply $2$ by $2$ .

$lim_{t \to - 6} \frac{2 t}{4 t + \frac{d}{d t} [13 t] + \frac{d}{d t} [6]}$

Step 1.3.8

Evaluate $\frac{d}{d t} [13 t]$ .

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Step 1.3.8.1

Since $13$ is constant with respect to $t$ , the derivative of $13 t$ with respect to $t$ is $13 \frac{d}{d t} [t]$ .

$lim_{t \to - 6} \frac{2 t}{4 t + 13 \frac{d}{d t} [t] + \frac{d}{d t} [6]}$

Step 1.3.8.2

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

$lim_{t \to - 6} \frac{2 t}{4 t + 13 \cdot 1 + \frac{d}{d t} [6]}$

Step 1.3.8.3

Multiply $13$ by $1$ .

$lim_{t \to - 6} \frac{2 t}{4 t + 13 + \frac{d}{d t} [6]}$

Step 1.3.9

Since $6$ is constant with respect to $t$ , the derivative of $6$ with respect to $t$ is $0$ .

$lim_{t \to - 6} \frac{2 t}{4 t + 13 + 0}$

Step 1.3.10

Add $4 t + 13$ and $0$ .

$lim_{t \to - 6} \frac{2 t}{4 t + 13}$

Step 2

Evaluate the limit.

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Step 2.1

Move the term $2$ outside of the limit because it is constant with respect to $t$ .

$2 lim_{t \to - 6} \frac{t}{4 t + 13}$

Step 2.2

Split the limit using the Limits Quotient Rule on the limit as $t$ approaches $- 6$ .

$2 \frac{lim_{t \to - 6} t}{lim_{t \to - 6} 4 t + 13}$

Step 2.3

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $- 6$ .

$2 \frac{lim_{t \to - 6} t}{lim_{t \to - 6} 4 t + lim_{t \to - 6} 13}$

Step 2.4

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

$2 \frac{lim_{t \to - 6} t}{4 lim_{t \to - 6} t + lim_{t \to - 6} 13}$

Step 2.5

Evaluate the limit of $13$ which is constant as $t$ approaches $- 6$ .

$2 \frac{lim_{t \to - 6} t}{4 lim_{t \to - 6} t + 13}$

Step 3

Evaluate the limits by plugging in $- 6$ for all occurrences of $t$ .

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Step 3.1

Evaluate the limit of $t$ by plugging in $- 6$ for $t$ .

$2 \frac{- 6}{4 lim_{t \to - 6} t + 13}$

Step 3.2

Evaluate the limit of $t$ by plugging in $- 6$ for $t$ .

$2 \frac{- 6}{4 \cdot - 6 + 13}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the denominator.

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Step 4.1.1

Multiply $4$ by $- 6$ .

$2 \frac{- 6}{- 24 + 13}$

Step 4.1.2

Add $- 24$ and $13$ .

$2 (\frac{- 6}{- 11})$

Step 4.2

Dividing two negative values results in a positive value.

$2 (\frac{6}{11})$

Step 4.3

Multiply $2 (\frac{6}{11})$ .

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Step 4.3.1

Combine $2$ and $\frac{6}{11}$ .

$\frac{2 \cdot 6}{11}$

Step 4.3.2

Multiply $2$ by $6$ .

$\frac{12}{11}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{12}{11}$

Decimal Form:

$1.\overline{09}$