Calculus Examples

$lim_{t \to 0} \frac{\sin (3 t)}{t}$

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.

$\frac{lim_{t \to 0} \sin (3 t)}{lim_{t \to 0} t}$

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

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

$\frac{\sin (lim_{t \to 0} 3 t)}{lim_{t \to 0} t}$

Step 1.1.2.1.2

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

$\frac{\sin (3 lim_{t \to 0} t)}{lim_{t \to 0} t}$

Step 1.1.2.2

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

$\frac{\sin (3 \cdot 0)}{lim_{t \to 0} t}$

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Multiply $3$ by $0$ .

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

Step 1.1.2.3.2

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

$\frac{0}{lim_{t \to 0} t}$

Step 1.1.3

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

$\frac{0}{0}$

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.

$lim_{t \to 0} \frac{\sin (3 t)}{t} = lim_{t \to 0} \frac{\frac{d}{d t} [\sin (3 t)]}{\frac{d}{d t} [t]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{t \to 0} \frac{\frac{d}{d t} [\sin (3 t)]}{\frac{d}{d t} [t]}$

Step 1.3.2

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 $g (t) = 3 t$ .

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Step 1.3.2.1

To apply the Chain Rule, set $u$ as $3 t$ .

$lim_{t \to 0} \frac{\frac{d}{d u} [\sin (u)] \frac{d}{d t} [3 t]}{\frac{d}{d t} [t]}$

Step 1.3.2.2

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

$lim_{t \to 0} \frac{\cos (u) \frac{d}{d t} [3 t]}{\frac{d}{d t} [t]}$

Step 1.3.2.3

Replace all occurrences of $u$ with $3 t$ .

$lim_{t \to 0} \frac{\cos (3 t) \frac{d}{d t} [3 t]}{\frac{d}{d t} [t]}$

Step 1.3.3

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

$lim_{t \to 0} \frac{\cos (3 t) (3 \frac{d}{d t} [t])}{\frac{d}{d t} [t]}$

Step 1.3.4

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

$lim_{t \to 0} \frac{\cos (3 t) (3 \cdot 1)}{\frac{d}{d t} [t]}$

Step 1.3.5

Multiply $3$ by $1$ .

$lim_{t \to 0} \frac{\cos (3 t) \cdot 3}{\frac{d}{d t} [t]}$

Step 1.3.6

Move $3$ to the left of $\cos (3 t)$ .

$lim_{t \to 0} \frac{3 \cdot \cos (3 t)}{\frac{d}{d t} [t]}$

Step 1.3.7

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

$lim_{t \to 0} \frac{3 \cos (3 t)}{1}$

Step 1.4

Divide $3 \cos (3 t)$ by $1$ .

$lim_{t \to 0} 3 \cos (3 t)$

Step 2

Evaluate the limit.

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Step 2.1

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

$3 lim_{t \to 0} \cos (3 t)$

Step 2.2

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

$3 \cos (lim_{t \to 0} 3 t)$

Step 2.3

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

$3 \cos (3 lim_{t \to 0} t)$

Step 3

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

$3 \cos (3 \cdot 0)$

Step 4

Simplify the answer.

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Step 4.1

Multiply $3$ by $0$ .

$3 \cos (0)$

Step 4.2

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

$3 \cdot 1$

Step 4.3

Multiply $3$ by $1$ .

$3$