Calculus Examples

$lim_{x \to 0} \frac{x}{\arctan (4 x)}$

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.

$\frac{lim_{x \to 0} x}{lim_{x \to 0} \arctan (4 x)}$

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

$\frac{0}{lim_{x \to 0} \arctan (4 x)}$

Evaluate the limit of the denominator.

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Multiply $4$ by $0$ .

$\frac{0}{0}$

Substitute $t$ for $4 x$ and let $t$ approach $0$ since $lim_{x \to 0} 4 x = 0$ .

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

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

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Multiply $4$ by $0$ .

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

The exact value of $\arctan (0)$ 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

$\frac{0}{0}$

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.

$lim_{x \to 0} \frac{x}{\arctan (4 x)} = lim_{x \to 0} \frac{\frac{d}{d x} [x]}{\frac{d}{d x} [\arctan (4 x)]}$

Find the derivative of the numerator and denominator.

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Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [x]}{\frac{d}{d x} [\arctan (4 x)]}$

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

$lim_{x \to 0} \frac{1}{\frac{d}{d x} [\arctan (4 x)]}$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \arctan (x)$ and $g (x) = 4 x$ .

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To apply the Chain Rule, set $u$ as $4 x$ .

$lim_{x \to 0} \frac{1}{\frac{d}{d u} [\arctan (u)] \frac{d}{d x} [4 x]}$

The derivative of $\arctan (u)$ with respect to $u$ is $\frac{1}{1 + u^{2}}$ .

$lim_{x \to 0} \frac{1}{\frac{1}{1 + u^{2}} \frac{d}{d x} [4 x]}$

Replace all occurrences of $u$ with $4 x$ .

$lim_{x \to 0} \frac{1}{\frac{1}{1 + {(4 x)}^{2}} \frac{d}{d x} [4 x]}$

Factor $4$ out of $4 x$ .

$lim_{x \to 0} \frac{1}{\frac{1}{1 + {(4 (x))}^{2}} \frac{d}{d x} [4 x]}$

Apply the product rule to $4 (x)$ .

$lim_{x \to 0} \frac{1}{\frac{1}{1 + 4^{2} x^{2}} \frac{d}{d x} [4 x]}$

Raise $4$ to the power of $2$ .

$lim_{x \to 0} \frac{1}{\frac{1}{1 + 16 x^{2}} \frac{d}{d x} [4 x]}$

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{1}{\frac{1}{1 + 16 x^{2}} (4 \frac{d}{d x} [x])}$

Combine $4$ and $\frac{1}{1 + 16 x^{2}}$ .

$lim_{x \to 0} \frac{1}{\frac{4}{1 + 16 x^{2}} \frac{d}{d x} [x]}$

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

$lim_{x \to 0} \frac{1}{\frac{4}{1 + 16 x^{2}} \cdot 1}$

Multiply $\frac{4}{1 + 16 x^{2}}$ by $1$ .

$lim_{x \to 0} \frac{1}{\frac{4}{1 + 16 x^{2}}}$

Reorder terms.

$lim_{x \to 0} \frac{1}{\frac{4}{16 x^{2} + 1}}$

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 0} 1 \frac{16 x^{2} + 1}{4}$

Evaluate the limit.

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Multiply $\frac{16 x^{2} + 1}{4}$ by $1$ .

$lim_{x \to 0} \frac{16 x^{2} + 1}{4}$

Move the term $\frac{1}{4}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{4} lim_{x \to 0} 16 x^{2} + 1$

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

$\frac{1}{4} (lim_{x \to 0} 16 x^{2} + lim_{x \to 0} 1)$

Move the term $16$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{4} (16 lim_{x \to 0} x^{2} + lim_{x \to 0} 1)$

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

$\frac{1}{4} (16 {(lim_{x \to 0} x)}^{2} + lim_{x \to 0} 1)$

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

$\frac{1}{4} (16 {(lim_{x \to 0} x)}^{2} + 1)$

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

$\frac{1}{4} (16 \cdot 0^{2} + 1)$

Simplify the answer.

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Simplify each term.

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

$\frac{1}{4} (16 \cdot 0 + 1)$

Multiply $16$ by $0$ .

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

Add $0$ and $1$ .

$\frac{1}{4} \cdot 1$

Multiply $\frac{1}{4}$ by $1$ .

$\frac{1}{4}$