Simplify limit as x approaches 0 of ((x+2)^5-32)/x

$lim_{x \to 0} \frac{{(x + 2)}^{5} - 32}{x}$

Apply L\'Hospital\'s rule.

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

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

$\frac{lim_{x \to 0} {(x + 2)}^{5} - 32}{lim_{x \to 0} x}$

Evaluate the limit of the numerator.

Evaluate the limit.

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

$\frac{lim_{x \to 0} {(x + 2)}^{5} - lim_{x \to 0} 32}{lim_{x \to 0} x}$

Move the exponent $5$ from ${(x + 2)}^{5}$ outside the limit using the Limits Power Rule.

$\frac{{(lim_{x \to 0} x + 2)}^{5} - lim_{x \to 0} 32}{lim_{x \to 0} x}$

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

$\frac{{(lim_{x \to 0} x + lim_{x \to 0} 2)}^{5} - lim_{x \to 0} 32}{lim_{x \to 0} x}$

Evaluate the limit of $2$ which is constant as $x$ approaches $0$ .

$\frac{{(lim_{x \to 0} x + 2)}^{5} - lim_{x \to 0} 32}{lim_{x \to 0} x}$

Evaluate the limit of $32$ which is constant as $x$ approaches $0$ .

$\frac{{(lim_{x \to 0} x + 2)}^{5} - 1 \cdot 32}{lim_{x \to 0} x}$

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

$\frac{{(0 + 2)}^{5} - 1 \cdot 32}{lim_{x \to 0} x}$

Simplify the answer.

Simplify each term.

Add $0$ and $2$ .

$\frac{2^{5} - 1 \cdot 32}{lim_{x \to 0} x}$

Raise $2$ to the power of $5$ .

$\frac{32 - 1 \cdot 32}{lim_{x \to 0} x}$

Multiply $- 1$ by $32$ .

$\frac{32 - 32}{lim_{x \to 0} x}$

Subtract $32$ from $32$ .

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

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

$\frac{0}{0}$

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.

$lim_{x \to 0} \frac{{(x + 2)}^{5} - 32}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [{(x + 2)}^{5} - 32]}{\frac{d}{d x} [x]}$

Find the derivative of the numerator and denominator.

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [{(x + 2)}^{5} - 32]}{\frac{d}{d x} [x]}$

By the Sum Rule, the derivative of ${(x + 2)}^{5} - 32$ with respect to $x$ is $\frac{d}{d x} [{(x + 2)}^{5}] + \frac{d}{d x} [- 32]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [{(x + 2)}^{5}] + \frac{d}{d x} [- 32]}{\frac{d}{d x} [x]}$

Evaluate $\frac{d}{d x} [{(x + 2)}^{5}]$ .

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{5}$ and $g (x) = x + 2$ .

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

$lim_{x \to 0} \frac{\frac{d}{d u} [u^{5}] \frac{d}{d x} [x + 2] + \frac{d}{d x} [- 32]}{\frac{d}{d x} [x]}$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 5$ .

$lim_{x \to 0} \frac{5 u^{4} \frac{d}{d x} [x + 2] + \frac{d}{d x} [- 32]}{\frac{d}{d x} [x]}$

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

$lim_{x \to 0} \frac{5 {(x + 2)}^{4} \frac{d}{d x} [x + 2] + \frac{d}{d x} [- 32]}{\frac{d}{d x} [x]}$

By the Sum Rule, the derivative of $x + 2$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [2]$ .

$lim_{x \to 0} \frac{5 {(x + 2)}^{4} (\frac{d}{d x} [x] + \frac{d}{d x} [2]) + \frac{d}{d x} [- 32]}{\frac{d}{d x} [x]}$

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

$lim_{x \to 0} \frac{5 {(x + 2)}^{4} (1 + \frac{d}{d x} [2]) + \frac{d}{d x} [- 32]}{\frac{d}{d x} [x]}$

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$lim_{x \to 0} \frac{5 {(x + 2)}^{4} (1 + 0) + \frac{d}{d x} [- 32]}{\frac{d}{d x} [x]}$

Add $1$ and $0$ .

$lim_{x \to 0} \frac{5 {(x + 2)}^{4} \cdot 1 + \frac{d}{d x} [- 32]}{\frac{d}{d x} [x]}$

Multiply $5$ by $1$ .

$lim_{x \to 0} \frac{5 {(x + 2)}^{4} + \frac{d}{d x} [- 32]}{\frac{d}{d x} [x]}$

Since $- 32$ is constant with respect to $x$ , the derivative of $- 32$ with respect to $x$ is $0$ .

$lim_{x \to 0} \frac{5 {(x + 2)}^{4} + 0}{\frac{d}{d x} [x]}$

Add $5 {(x + 2)}^{4}$ and $0$ .

$lim_{x \to 0} \frac{5 {(x + 2)}^{4}}{\frac{d}{d x} [x]}$

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

$lim_{x \to 0} \frac{5 {(x + 2)}^{4}}{1}$

Divide $5 {(x + 2)}^{4}$ by $1$ .

$lim_{x \to 0} 5 {(x + 2)}^{4}$

Evaluate the limit.

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

$5 lim_{x \to 0} {(x + 2)}^{4}$

Move the exponent $4$ from ${(x + 2)}^{4}$ outside the limit using the Limits Power Rule.

$5 {(lim_{x \to 0} x + 2)}^{4}$

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

$5 {(lim_{x \to 0} x + lim_{x \to 0} 2)}^{4}$

Evaluate the limit of $2$ which is constant as $x$ approaches $0$ .

$5 {(lim_{x \to 0} x + 2)}^{4}$

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

$5 {(0 + 2)}^{4}$

Simplify the expression.

Add $0$ and $2$ .

$5 \cdot 2^{4}$

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

$5 \cdot 16$

Multiply $5$ by $16$ .

$80$

Simplify limit as x approaches 0 of ((x+2)^5-32)/x

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