Calculus Examples

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

Step 1

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

Tap for more steps...

Step 1.1

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

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

Step 1.2

Evaluate the limit of the numerator.

Tap for more steps...

Step 1.2.1

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

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

Step 1.2.2

Move the limit into the exponent.

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

Step 1.2.3

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

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

Step 1.2.4

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

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

Step 1.2.5

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

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

Step 1.2.6

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

Tap for more steps...

Step 1.2.6.1

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

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

Step 1.2.6.2

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

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

Step 1.2.7

Simplify the answer.

Tap for more steps...

Step 1.2.7.1

Simplify each term.

Tap for more steps...

Step 1.2.7.1.1

Multiply $5$ by $0$ .

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

Step 1.2.7.1.2

Anything raised to $0$ is $1$ .

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

Step 1.2.7.1.3

Multiply $- 1$ by $1$ .

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

Step 1.2.7.1.4

Multiply $- 5$ by $0$ .

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

Step 1.2.7.2

Subtract $1$ from $1$ .

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

Step 1.2.7.3

Add $0$ and $0$ .

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

Step 1.3

Evaluate the limit of the denominator.

Tap for more steps...

Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

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

$\frac{0}{0}$

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.

$lim_{x \to 0} \frac{e^{5 x} - 1 - 5 x}{x^{2}} = lim_{x \to 0} \frac{\frac{d}{d x} [e^{5 x} - 1 - 5 x]}{\frac{d}{d x} [x^{2}]}$

Step 3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [e^{5 x} - 1 - 5 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.2

By the Sum Rule, the derivative of $e^{5 x} - 1 - 5 x$ with respect to $x$ is $\frac{d}{d x} [e^{5 x}] + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 5 x]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [e^{5 x}] + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 5 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3

Evaluate $\frac{d}{d x} [e^{5 x}]$ .

Tap for more steps...

Step 3.3.1

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

Tap for more steps...

Step 3.3.1.1

To apply the Chain Rule, set $u$ as $5 x$ .

$lim_{x \to 0} \frac{\frac{d}{d u} [e^{u}] \frac{d}{d x} [5 x] + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 5 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$lim_{x \to 0} \frac{e^{u} \frac{d}{d x} [5 x] + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 5 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3.1.3

Replace all occurrences of $u$ with $5 x$ .

$lim_{x \to 0} \frac{e^{5 x} \frac{d}{d x} [5 x] + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 5 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3.2

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

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

Step 3.3.3

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

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

Step 3.3.4

Multiply $5$ by $1$ .

$lim_{x \to 0} \frac{e^{5 x} \cdot 5 + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 5 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3.5

Move $5$ to the left of $e^{5 x}$ .

$lim_{x \to 0} \frac{5 e^{5 x} + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 5 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

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

Step 3.5

Evaluate $\frac{d}{d x} [- 5 x]$ .

Tap for more steps...

Step 3.5.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

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

Step 3.5.2

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

$lim_{x \to 0} \frac{5 e^{5 x} + 0 - 5 \cdot 1}{\frac{d}{d x} [x^{2}]}$

Step 3.5.3

Multiply $- 5$ by $1$ .

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

Step 3.6

Add $5 e^{5 x}$ and $0$ .

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

Step 3.7

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

$lim_{x \to 0} \frac{5 e^{5 x} - 5}{2 x}$

Step 4

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

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

Step 5

Apply L'Hospital's rule.

Tap for more steps...

Step 5.1

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

Tap for more steps...

Step 5.1.1

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

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

Step 5.1.2

Evaluate the limit of the numerator.

Tap for more steps...

Step 5.1.2.1

Evaluate the limit.

Tap for more steps...

Step 5.1.2.1.1

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

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

Step 5.1.2.1.2

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

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

Step 5.1.2.1.3

Move the limit into the exponent.

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

Step 5.1.2.1.4

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

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

Step 5.1.2.1.5

Evaluate the limit of $5$ which is constant as $x$ approaches $0$ .

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

Step 5.1.2.2

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

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

Step 5.1.2.3

Simplify the answer.

Tap for more steps...

Step 5.1.2.3.1

Simplify each term.

Tap for more steps...

Step 5.1.2.3.1.1

Multiply $5$ by $0$ .

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

Step 5.1.2.3.1.2

Anything raised to $0$ is $1$ .

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

Step 5.1.2.3.1.3

Multiply $5$ by $1$ .

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

Step 5.1.2.3.1.4

Multiply $- 1$ by $5$ .

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

Step 5.1.2.3.2

Subtract $5$ from $5$ .

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

Step 5.1.3

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

$\frac{1}{2} \cdot \frac{0}{0}$

Step 5.1.4

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

Undefined

$\frac{1}{2} \cdot \frac{0}{0}$

Step 5.2

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.

$lim_{x \to 0} \frac{5 e^{5 x} - 5}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [5 e^{5 x} - 5]}{\frac{d}{d x} [x]}$

Step 5.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 5.3.1

Differentiate the numerator and denominator.

$\frac{1}{2} lim_{x \to 0} \frac{\frac{d}{d x} [5 e^{5 x} - 5]}{\frac{d}{d x} [x]}$

Step 5.3.2

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

$\frac{1}{2} lim_{x \to 0} \frac{\frac{d}{d x} [5 e^{5 x}] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x]}$

Step 5.3.3

Evaluate $\frac{d}{d x} [5 e^{5 x}]$ .

Tap for more steps...

Step 5.3.3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{5 x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{5 x}]$ .

$\frac{1}{2} lim_{x \to 0} \frac{5 \frac{d}{d x} [e^{5 x}] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x]}$

Step 5.3.3.2

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

Tap for more steps...

Step 5.3.3.2.1

To apply the Chain Rule, set $u$ as $5 x$ .

$\frac{1}{2} lim_{x \to 0} \frac{5 \frac{d}{d u} [e^{u}] \frac{d}{d x} [5 x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x]}$

Step 5.3.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{1}{2} lim_{x \to 0} \frac{5 e^{u} \frac{d}{d x} [5 x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x]}$

Step 5.3.3.2.3

Replace all occurrences of $u$ with $5 x$ .

$\frac{1}{2} lim_{x \to 0} \frac{5 e^{5 x} \frac{d}{d x} [5 x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x]}$

Step 5.3.3.3

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$\frac{1}{2} lim_{x \to 0} \frac{5 e^{5 x} \cdot 5 \frac{d}{d x} [x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x]}$

Step 5.3.3.4

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

$\frac{1}{2} lim_{x \to 0} \frac{5 e^{5 x} \cdot 5 \cdot 1 + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x]}$

Step 5.3.3.5

Multiply $5$ by $1$ .

$\frac{1}{2} lim_{x \to 0} \frac{5 e^{5 x} \cdot 5 + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x]}$

Step 5.3.3.6

Move $5$ to the left of $e^{5 x}$ .

$\frac{1}{2} lim_{x \to 0} \frac{5 \cdot 5 \cdot e^{5 x} + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x]}$

Step 5.3.3.7

Multiply $5$ by $5$ .

$\frac{1}{2} lim_{x \to 0} \frac{25 e^{5 x} + \frac{d}{d x} [- 5]}{\frac{d}{d x} [x]}$

Step 5.3.4

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$\frac{1}{2} lim_{x \to 0} \frac{25 e^{5 x} + 0}{\frac{d}{d x} [x]}$

Step 5.3.5

Add $25 e^{5 x}$ and $0$ .

$\frac{1}{2} lim_{x \to 0} \frac{25 e^{5 x}}{\frac{d}{d x} [x]}$

Step 5.3.6

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

$\frac{1}{2} lim_{x \to 0} \frac{25 e^{5 x}}{1}$

Step 5.4

Divide $25 e^{5 x}$ by $1$ .

$\frac{1}{2} lim_{x \to 0} 25 e^{5 x}$

Step 6

Evaluate the limit.

Tap for more steps...

Step 6.1

Move the term $25$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} \cdot 25 lim_{x \to 0} e^{5 x}$

Step 6.2

Move the limit into the exponent.

$\frac{1}{2} \cdot 25 e^{lim_{x \to 0} 5 x}$

Step 6.3

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

$\frac{1}{2} \cdot 25 e^{5 lim_{x \to 0} x}$

Step 7

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

$\frac{1}{2} \cdot 25 e^{5 \cdot 0}$

Step 8

Simplify the answer.

Tap for more steps...

Step 8.1

Combine $\frac{1}{2}$ and $25$ .

$\frac{25}{2} e^{5 \cdot 0}$

Step 8.2

Multiply $5$ by $0$ .

$\frac{25}{2} e^{0}$

Step 8.3

Anything raised to $0$ is $1$ .

$\frac{25}{2} \cdot 1$

Step 8.4

Multiply $\frac{25}{2}$ by $1$ .

$\frac{25}{2}$