Calculus Examples

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

Step 1

Rewrite $x \sin (\frac{1}{x})$ as $\frac{\sin (\frac{1}{x})}{\frac{1}{x}}$ .

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

Step 2

Apply L'Hospital's rule.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

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

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

Step 2.1.2.2

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x}$ approaches $0$ .

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

Step 2.1.2.3

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

$\frac{0}{lim_{x \to \infty} \frac{1}{x}}$

Step 2.1.3

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x}$ approaches $0$ .

$\frac{0}{0}$

Step 2.1.4

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

Undefined

$\frac{0}{0}$

Step 2.2

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.

$lim_{x \to \infty} \frac{\sin (\frac{1}{x})}{\frac{1}{x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [\sin (\frac{1}{x})]}{\frac{d}{d x} [\frac{1}{x}]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [\sin (\frac{1}{x})]}{\frac{d}{d x} [\frac{1}{x}]}$

Step 2.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = \frac{1}{x}$ .

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Step 2.3.2.1

To apply the Chain Rule, set $u$ as $\frac{1}{x}$ .

$lim_{x \to \infty} \frac{\frac{d}{d u} [\sin (u)] \frac{d}{d x} [\frac{1}{x}]}{\frac{d}{d x} [\frac{1}{x}]}$

Step 2.3.2.2

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

$lim_{x \to \infty} \frac{\cos (u) \frac{d}{d x} [\frac{1}{x}]}{\frac{d}{d x} [\frac{1}{x}]}$

Step 2.3.2.3

Replace all occurrences of $u$ with $\frac{1}{x}$ .

$lim_{x \to \infty} \frac{\cos (\frac{1}{x}) \frac{d}{d x} [\frac{1}{x}]}{\frac{d}{d x} [\frac{1}{x}]}$

Step 2.3.3

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$lim_{x \to \infty} \frac{\cos (\frac{1}{x}) \frac{d}{d x} [x^{- 1}]}{\frac{d}{d x} [\frac{1}{x}]}$

Step 2.3.4

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

$lim_{x \to \infty} \frac{\cos (\frac{1}{x}) (- x^{- 2})}{\frac{d}{d x} [\frac{1}{x}]}$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to \infty} \frac{\cos (\frac{1}{x}) (- \frac{1}{x^{2}})}{\frac{d}{d x} [\frac{1}{x}]}$

Step 2.3.5.2

Combine $\cos (\frac{1}{x})$ and $\frac{1}{x^{2}}$ .

$lim_{x \to \infty} \frac{- \frac{\cos (\frac{1}{x})}{x^{2}}}{\frac{d}{d x} [\frac{1}{x}]}$

Step 2.3.6

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$lim_{x \to \infty} \frac{- \frac{\cos (\frac{1}{x})}{x^{2}}}{\frac{d}{d x} [x^{- 1}]}$

Step 2.3.7

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

$lim_{x \to \infty} \frac{- \frac{\cos (\frac{1}{x})}{x^{2}}}{- x^{- 2}}$

Step 2.3.8

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to \infty} \frac{- \frac{\cos (\frac{1}{x})}{x^{2}}}{- \frac{1}{x^{2}}}$

Step 2.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to \infty} - \frac{\cos (\frac{1}{x})}{x^{2}} (- x^{2})$

Step 2.5

Combine factors.

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Step 2.5.1

Multiply $- 1$ by $- 1$ .

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

Step 2.5.2

Multiply $\frac{\cos (\frac{1}{x})}{x^{2}}$ by $1$ .

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

Step 2.5.3

Combine $\frac{\cos (\frac{1}{x})}{x^{2}}$ and $x^{2}$ .

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

Step 2.6

Cancel the common factor of $x^{2}$ .

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Step 2.6.1

Cancel the common factor.

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

Step 2.6.2

Divide $\cos (\frac{1}{x})$ by $1$ .

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

Step 3

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

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

Step 4

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x}$ approaches $0$ .

$\cos (0)$

Step 5

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

$1$