Calculus Examples

$lim_{t \to 0} \frac{\tan (8 t)}{\sin (2 t)}$

Step 1

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

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

$\frac{lim_{t \to 0} \tan (8 t)}{lim_{t \to 0} \sin (2 t)}$

Evaluate the limit of the numerator.

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Evaluate the limit.

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Move the limit inside the trig function because tangent is continuous.

$\frac{\tan (lim_{t \to 0} 8 t)}{lim_{t \to 0} \sin (2 t)}$

Move the term $8$ outside of the limit because it is constant with respect to $t$ .

$\frac{\tan (8 lim_{t \to 0} t)}{lim_{t \to 0} \sin (2 t)}$

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

$\frac{\tan (8 \cdot 0)}{lim_{t \to 0} \sin (2 t)}$

Simplify the answer.

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Multiply $8$ by $0$ .

$\frac{\tan (0)}{lim_{t \to 0} \sin (2 t)}$

The exact value of $\tan (0)$ is $0$ .

$\frac{0}{lim_{t \to 0} \sin (2 t)}$

Evaluate the limit of the denominator.

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Evaluate the limit.

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Move the limit inside the trig function because sine is continuous.

$\frac{0}{\sin (lim_{t \to 0} 2 t)}$

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

$\frac{0}{\sin (2 lim_{t \to 0} t)}$

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

$\frac{0}{\sin (2 \cdot 0)}$

Simplify the answer.

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Multiply $2$ by $0$ .

$\frac{0}{\sin (0)}$

The exact value of $\sin (0)$ is $0$ .

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

Step 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.

$lim_{t \to 0} \frac{\tan (8 t)}{\sin (2 t)} = lim_{t \to 0} \frac{\frac{d}{d t} [\tan (8 t)]}{\frac{d}{d t} [\sin (2 t)]}$

Step 3

Find the derivative of the numerator and denominator.

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Differentiate the numerator and denominator.

$lim_{t \to 0} \frac{\frac{d}{d t} [\tan (8 t)]}{\frac{d}{d t} [\sin (2 t)]}$

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \tan (t)$ and $g (t) = 8 t$ .

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To apply the Chain Rule, set $u$ as $8 t$ .

$lim_{t \to 0} \frac{\frac{d}{d u} [\tan (u)] \frac{d}{d t} [8 t]}{\frac{d}{d t} [\sin (2 t)]}$

The derivative of $\tan (u)$ with respect to $u$ is $\sec^{2} (u)$ .

$lim_{t \to 0} \frac{\sec^{2} (u) \frac{d}{d t} [8 t]}{\frac{d}{d t} [\sin (2 t)]}$

Replace all occurrences of $u$ with $8 t$ .

$lim_{t \to 0} \frac{\sec^{2} (8 t) \frac{d}{d t} [8 t]}{\frac{d}{d t} [\sin (2 t)]}$

Since $8$ is constant with respect to $t$ , the derivative of $8 t$ with respect to $t$ is $8 \frac{d}{d t} [t]$ .

$lim_{t \to 0} \frac{\sec^{2} (8 t) (8 \frac{d}{d t} [t])}{\frac{d}{d t} [\sin (2 t)]}$

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

$lim_{t \to 0} \frac{\sec^{2} (8 t) (8 \cdot 1)}{\frac{d}{d t} [\sin (2 t)]}$

Multiply $8$ by $1$ .

$lim_{t \to 0} \frac{\sec^{2} (8 t) \cdot 8}{\frac{d}{d t} [\sin (2 t)]}$

Move $8$ to the left of $\sec^{2} (8 t)$ .

$lim_{t \to 0} \frac{8 \cdot \sec^{2} (8 t)}{\frac{d}{d t} [\sin (2 t)]}$

Multiply $8$ by $\sec^{2} (8 t)$ .

$lim_{t \to 0} \frac{8 \sec^{2} (8 t)}{\frac{d}{d t} [\sin (2 t)]}$

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

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To apply the Chain Rule, set $u$ as $2 t$ .

$lim_{t \to 0} \frac{8 \sec^{2} (8 t)}{\frac{d}{d u} [\sin (u)] \frac{d}{d t} [2 t]}$

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$lim_{t \to 0} \frac{8 \sec^{2} (8 t)}{\cos (u) \frac{d}{d t} [2 t]}$

Replace all occurrences of $u$ with $2 t$ .

$lim_{t \to 0} \frac{8 \sec^{2} (8 t)}{\cos (2 t) \frac{d}{d t} [2 t]}$

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

$lim_{t \to 0} \frac{8 \sec^{2} (8 t)}{\cos (2 t) (2 \frac{d}{d t} [t])}$

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

$lim_{t \to 0} \frac{8 \sec^{2} (8 t)}{\cos (2 t) (2 \cdot 1)}$

Multiply $2$ by $1$ .

$lim_{t \to 0} \frac{8 \sec^{2} (8 t)}{\cos (2 t) \cdot 2}$

Move $2$ to the left of $\cos (2 t)$ .

$lim_{t \to 0} \frac{8 \sec^{2} (8 t)}{2 \cdot \cos (2 t)}$

Multiply $2$ by $\cos (2 t)$ .

$lim_{t \to 0} \frac{8 \sec^{2} (8 t)}{2 \cos (2 t)}$

Step 4

Cancel the common factor of $8$ and $2$ .

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Factor $2$ out of $8 \sec^{2} (8 t)$ .

$lim_{t \to 0} \frac{2 (4 \sec^{2} (8 t))}{2 \cos (2 t)}$

Cancel the common factors.

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Factor $2$ out of $2 \cos (2 t)$ .

$lim_{t \to 0} \frac{2 (4 \sec^{2} (8 t))}{2 (\cos (2 t))}$

Cancel the common factor.

$lim_{t \to 0} \frac{2 (4 \sec^{2} (8 t))}{2 \cos (2 t)}$

Rewrite the expression.

$lim_{t \to 0} \frac{4 \sec^{2} (8 t)}{\cos (2 t)}$

Step 5

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

$4 lim_{t \to 0} \frac{\sec^{2} (8 t)}{\cos (2 t)}$

Step 6

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

$4 \frac{lim_{t \to 0} \sec^{2} (8 t)}{lim_{t \to 0} \cos (2 t)}$

Step 7

Move the exponent $2$ from $\sec^{2} (8 t)$ outside the limit using the Limits Power Rule.

$4 \frac{{(lim_{t \to 0} \sec (8 t))}^{2}}{lim_{t \to 0} \cos (2 t)}$

Step 8

Move the limit inside the trig function because secant is continuous.

$4 \frac{\sec^{2} (lim_{t \to 0} 8 t)}{lim_{t \to 0} \cos (2 t)}$

Step 9

Move the term $8$ outside of the limit because it is constant with respect to $t$ .

$4 \frac{\sec^{2} (8 lim_{t \to 0} t)}{lim_{t \to 0} \cos (2 t)}$

Step 10

Move the limit inside the trig function because cosine is continuous.

$4 \frac{\sec^{2} (8 lim_{t \to 0} t)}{\cos (lim_{t \to 0} 2 t)}$

Step 11

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

$4 \frac{\sec^{2} (8 lim_{t \to 0} t)}{\cos (2 lim_{t \to 0} t)}$

Step 12

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

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

$4 \frac{\sec^{2} (8 \cdot 0)}{\cos (2 lim_{t \to 0} t)}$

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

$4 \frac{\sec^{2} (8 \cdot 0)}{\cos (2 \cdot 0)}$

Step 13

Simplify the answer.

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Simplify the numerator.

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Multiply $8$ by $0$ .

$4 \frac{\sec^{2} (0)}{\cos (2 \cdot 0)}$

The exact value of $\sec (0)$ is $1$ .

$4 \frac{1^{2}}{\cos (2 \cdot 0)}$

One to any power is one.

$4 \frac{1}{\cos (2 \cdot 0)}$

Simplify the denominator.

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Multiply $2$ by $0$ .

$4 \frac{1}{\cos (0)}$

The exact value of $\cos (0)$ is $1$ .

$4 (\frac{1}{1})$

Cancel the common factor of $1$ .

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Cancel the common factor.

$4 (\frac{1}{1})$

Rewrite the expression.

$4 \cdot 1$

Multiply $4$ by $1$ .

$4$