Calculus Examples

$lim_{x \to \infty} \frac{e^{x} + x^{2}}{e^{x} - x}$

Step 1

Divide the numerator and denominator by the fastest growing term in the denominator.

$lim_{x \to \infty} \frac{\frac{e^{x}}{e^{x}} + \frac{x^{2}}{e^{x}}}{\frac{e^{x}}{e^{x}} - \frac{x}{e^{x}}}$

Step 2

Evaluate the limit.

Tap for more steps...

Step 2.1

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

Tap for more steps...

Step 2.1.1

Cancel the common factor.

$lim_{x \to \infty} \frac{\frac{e^{x}}{e^{x}} + \frac{x^{2}}{e^{x}}}{\frac{e^{x}}{e^{x}} - \frac{x}{e^{x}}}$

Step 2.1.2

Rewrite the expression.

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

Step 2.2

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

Tap for more steps...

Step 2.2.1

Cancel the common factor.

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

Step 2.2.2

Rewrite the expression.

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

Step 2.3

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $\infty$ .

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

Step 2.4

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

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

Step 2.5

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

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

Step 3

Apply L'Hospital's rule.

Tap for more steps...

Step 3.1

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

Tap for more steps...

Step 3.1.1

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

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

Step 3.1.2

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$\frac{1 + \frac{\infty}{lim_{x \to \infty} e^{x}}}{lim_{x \to \infty} 1 - \frac{x}{e^{x}}}$

Step 3.1.3

Since the exponent $x$ approaches $\infty$ , the quantity $e^{x}$ approaches $\infty$ .

$\frac{1 + \frac{\infty}{\infty}}{lim_{x \to \infty} 1 - \frac{x}{e^{x}}}$

Step 3.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{1 + \frac{\infty}{\infty}}{lim_{x \to \infty} 1 - \frac{x}{e^{x}}}$

Step 3.2

Since $\frac{\infty}{\infty}$ 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.

$lim_{x \to \infty} \frac{x^{2}}{e^{x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{x}]}$

Step 3.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 3.3.1

Differentiate the numerator and denominator.

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 4

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

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

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.

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

Step 5.1.2

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

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

Step 5.1.3

Since the exponent $x$ approaches $\infty$ , the quantity $e^{x}$ approaches $\infty$ .

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

Step 5.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 5.2

$lim_{x \to \infty} \frac{x}{e^{x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [x]}{\frac{d}{d x} [e^{x}]}$

Step 5.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 5.3.1

Differentiate the numerator and denominator.

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 6

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

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

Step 7

Evaluate the limit.

Tap for more steps...

Step 7.1

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

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

Step 7.2

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

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

Step 8

Apply L'Hospital's rule.

Tap for more steps...

Step 8.1

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

Tap for more steps...

Step 8.1.1

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

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

Step 8.1.2

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

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

Step 8.1.3

Since the exponent $x$ approaches $\infty$ , the quantity $e^{x}$ approaches $\infty$ .

$\frac{1 + 2 \cdot 0}{1 - \frac{\infty}{\infty}}$

Step 8.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{1 + 2 \cdot 0}{1 - \frac{\infty}{\infty}}$

Step 8.2

$lim_{x \to \infty} \frac{x}{e^{x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [x]}{\frac{d}{d x} [e^{x}]}$

Step 8.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 8.3.1

Differentiate the numerator and denominator.

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 9

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

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

Step 10

Simplify the answer.

Tap for more steps...

Step 10.1

Simplify the numerator.

Tap for more steps...

Step 10.1.1

Multiply $2$ by $0$ .

$\frac{1 + 0}{1 - 0}$

Step 10.1.2

Add $1$ and $0$ .

$\frac{1}{1 - 0}$

Step 10.2

Simplify the denominator.

Tap for more steps...

Step 10.2.1

Multiply $- 1$ by $0$ .

$\frac{1}{1 + 0}$

Step 10.2.2

Add $1$ and $0$ .

$\frac{1}{1}$

Step 10.3

Divide $1$ by $1$ .

$1$