Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Evaluate the limit.

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Step 1.3.1.1

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

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

Step 1.3.1.2

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

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

Step 1.3.2

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

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

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

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

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

Step 1.3.3.2

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

$\frac{0}{0}$

Step 1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 1.3.4

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

Undefined

$\frac{0}{0}$

Step 1.4

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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

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

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

Step 3.3

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 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.4

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

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

Step 3.5

Simplify.

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Step 3.5.1

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

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

Step 3.5.2

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

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

Step 3.5.3

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

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

Step 3.5.4

Apply the sine double-angle identity.

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

Step 4

The limit of $\frac{2 x}{\sin (2 x)}$ as $x$ approaches $0$ is $1$ .

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Step 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

Evaluate the limit of the numerator.

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Step 4.1.2.1

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

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

Step 4.1.2.2

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

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

Step 4.1.2.3

Multiply $2$ by $0$ .

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

Step 4.1.3

Evaluate the limit of the denominator.

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Step 4.1.3.1

Evaluate the limit.

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Step 4.1.3.1.1

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

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

Step 4.1.3.1.2

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

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

Step 4.1.3.2

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

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

Step 4.1.3.3

Simplify the answer.

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Step 4.1.3.3.1

Multiply $2$ by $0$ .

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

Step 4.1.3.3.2

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

$\frac{0}{0}$

Step 4.1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 4.1.3.4

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

Undefined

$\frac{0}{0}$

Step 4.1.4

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

Undefined

$\frac{0}{0}$

Step 4.2

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.

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

Step 4.3

Find the derivative of the numerator and denominator.

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Step 4.3.1

Differentiate the numerator and denominator.

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

Step 4.3.2

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

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

Step 4.3.3

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

$lim_{x \to 0} \frac{2 \cdot 1}{\frac{d}{d x} [\sin (2 x)]}$

Step 4.3.4

Multiply $2$ by $1$ .

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

Step 4.3.5

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

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Step 4.3.5.1

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

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

Step 4.3.5.2

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

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

Step 4.3.5.3

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

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

Step 4.3.6

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

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

Step 4.3.7

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

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

Step 4.3.8

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{2}{\cos (2 x) \cdot 2}$

Step 4.3.9

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

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

Step 4.4

Cancel the common factor of $2$ .

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Step 4.4.1

Cancel the common factor.

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

Step 4.4.2

Rewrite the expression.

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

Step 4.5

Convert from $\frac{1}{\cos (2 x)}$ to $\sec (2 x)$ .

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

Step 4.6

Evaluate the limit.

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Step 4.6.1

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

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

Step 4.6.2

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

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

Step 4.7

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

$\sec (2 \cdot 0)$

Step 4.8

Simplify the answer.

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Step 4.8.1

Multiply $2$ by $0$ .

$\sec (0)$

Step 4.8.2

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

$1$