Calculus Examples

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

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.

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

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 exponent $2$ from $\sin^{2} (x)$ outside the limit using the Limits Power Rule.

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

Step 1.1.2.1.2

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

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

$\frac{0^{2}}{lim_{x \to 0} x}$

Step 1.1.2.3.2

Raising $0$ to any positive power yields $0$ .

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

Step 1.1.3

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

$\frac{0}{0}$

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.

$lim_{x \to 0} \frac{\sin^{2} (x)}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [\sin^{2} (x)]}{\frac{d}{d x} [x]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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Step 1.3.2.1

To apply the Chain Rule, set $u$ as $\sin (x)$ .

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

Step 1.3.2.2

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

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

Step 1.3.2.3

Replace all occurrences of $u$ with $\sin (x)$ .

$lim_{x \to 0} \frac{2 \sin (x) \frac{d}{d x} [\sin (x)]}{\frac{d}{d x} [x]}$

Step 1.3.3

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

$lim_{x \to 0} \frac{2 \sin (x) \cos (x)}{\frac{d}{d x} [x]}$

Step 1.3.4

Simplify.

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Step 1.3.4.1

Reorder the factors of $2 \sin (x) \cos (x)$ .

$lim_{x \to 0} \frac{2 \cos (x) \sin (x)}{\frac{d}{d x} [x]}$

Step 1.3.4.2

Reorder $2 \cos (x)$ and $\sin (x)$ .

$lim_{x \to 0} \frac{\sin (x) (2 \cos (x))}{\frac{d}{d x} [x]}$

Step 1.3.4.3

Reorder $\sin (x)$ and $2$ .

$lim_{x \to 0} \frac{2 \cdot \sin (x) \cos (x)}{\frac{d}{d x} [x]}$

Step 1.3.4.4

Apply the sine double-angle identity.

$lim_{x \to 0} \frac{\sin (2 x)}{\frac{d}{d x} [x]}$

Step 1.3.5

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

$lim_{x \to 0} \frac{\sin (2 x)}{1}$

Step 1.4

Divide $\sin (2 x)$ by $1$ .

$lim_{x \to 0} \sin (2 x)$

Step 2

Evaluate the limit.

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Step 2.1

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

$\sin (lim_{x \to 0} 2 x)$

Step 2.2

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

$\sin (2 lim_{x \to 0} x)$

Step 3

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

$\sin (2 \cdot 0)$

Step 4

Simplify the answer.

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Step 4.1

Multiply $2$ by $0$ .

$\sin (0)$

Step 4.2

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

$0$