Calculus Examples

$lim_{x \to \infty} \frac{x^{3}}{e^{x^{2}}}$

Step 1

Apply L'Hospital's rule.

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Step 1.1

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

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Step 1.1.1

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

$\frac{lim_{x \to \infty} x^{3}}{lim_{x \to \infty} e^{x^{2}}}$

Step 1.1.2

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

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

Step 1.1.3

Since the exponent $x^{2}$ approaches $\infty$ , the quantity $e^{x^{2}}$ approaches $\infty$ .

$\frac{\infty}{\infty}$

Step 1.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{\infty}{\infty}$

Step 1.2

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.

$lim_{x \to \infty} \frac{x^{3}}{e^{x^{2}}} = lim_{x \to \infty} \frac{\frac{d}{d x} [x^{3}]}{\frac{d}{d x} [e^{x^{2}}]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [x^{3}]}{\frac{d}{d x} [e^{x^{2}}]}$

Step 1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$lim_{x \to \infty} \frac{3 x^{2}}{\frac{d}{d x} [e^{x^{2}}]}$

Step 1.3.3

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 $g (x) = x^{2}$ .

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Step 1.3.3.1

To apply the Chain Rule, set $u$ as $x^{2}$ .

$lim_{x \to \infty} \frac{3 x^{2}}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [x^{2}]}$

Step 1.3.3.2

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

$lim_{x \to \infty} \frac{3 x^{2}}{e^{u} \frac{d}{d x} [x^{2}]}$

Step 1.3.3.3

Replace all occurrences of $u$ with $x^{2}$ .

$lim_{x \to \infty} \frac{3 x^{2}}{e^{x^{2}} \frac{d}{d x} [x^{2}]}$

Step 1.3.4

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

$lim_{x \to \infty} \frac{3 x^{2}}{e^{x^{2}} (2 x)}$

Step 1.3.5

Simplify.

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Step 1.3.5.1

Reorder the factors of $e^{x^{2}} (2 x)$ .

$lim_{x \to \infty} \frac{3 x^{2}}{2 e^{x^{2}} x}$

Step 1.3.5.2

Reorder factors in $2 e^{x^{2}} x$ .

$lim_{x \to \infty} \frac{3 x^{2}}{2 x e^{x^{2}}}$

Step 1.4

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

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Step 1.4.1

Factor $x$ out of $3 x^{2}$ .

$lim_{x \to \infty} \frac{x (3 x)}{2 x e^{x^{2}}}$

Step 1.4.2

Cancel the common factors.

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Step 1.4.2.1

Factor $x$ out of $2 x e^{x^{2}}$ .

$lim_{x \to \infty} \frac{x (3 x)}{x (2 e^{x^{2}})}$

Step 1.4.2.2

Cancel the common factor.

$lim_{x \to \infty} \frac{x (3 x)}{x (2 e^{x^{2}})}$

Step 1.4.2.3

Rewrite the expression.

$lim_{x \to \infty} \frac{3 x}{2 e^{x^{2}}}$

Step 2

Move the term $\frac{3}{2}$ outside of the limit because it is constant with respect to $x$ .

$\frac{3}{2} lim_{x \to \infty} \frac{x}{e^{x^{2}}}$

Step 3

Apply L'Hospital's rule.

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Step 3.1

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

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Step 3.1.1

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

$\frac{3}{2} \cdot \frac{lim_{x \to \infty} x}{lim_{x \to \infty} e^{x^{2}}}$

Step 3.1.2

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

$\frac{3}{2} \cdot \frac{\infty}{lim_{x \to \infty} e^{x^{2}}}$

Step 3.1.3

Since the exponent $x^{2}$ approaches $\infty$ , the quantity $e^{x^{2}}$ approaches $\infty$ .

$\frac{3}{2} \cdot \frac{\infty}{\infty}$

Step 3.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{3}{2} \cdot \frac{\infty}{\infty}$

Step 3.2

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

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

$\frac{3}{2} lim_{x \to \infty} \frac{\frac{d}{d x} [x]}{\frac{d}{d x} [e^{x^{2}}]}$

Step 3.3.2

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

$\frac{3}{2} lim_{x \to \infty} \frac{1}{\frac{d}{d x} [e^{x^{2}}]}$

Step 3.3.3

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 $g (x) = x^{2}$ .

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Step 3.3.3.1

To apply the Chain Rule, set $u$ as $x^{2}$ .

$\frac{3}{2} lim_{x \to \infty} \frac{1}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [x^{2}]}$

Step 3.3.3.2

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

$\frac{3}{2} lim_{x \to \infty} \frac{1}{e^{u} \frac{d}{d x} [x^{2}]}$

Step 3.3.3.3

Replace all occurrences of $u$ with $x^{2}$ .

$\frac{3}{2} lim_{x \to \infty} \frac{1}{e^{x^{2}} \frac{d}{d x} [x^{2}]}$

Step 3.3.4

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

$\frac{3}{2} lim_{x \to \infty} \frac{1}{e^{x^{2}} \cdot 2 x}$

Step 3.3.5

Simplify.

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Step 3.3.5.1

Reorder the factors of $e^{x^{2}} (2 x)$ .

$\frac{3}{2} lim_{x \to \infty} \frac{1}{2 e^{x^{2}} x}$

Step 3.3.5.2

Reorder factors in $2 e^{x^{2}} x$ .

$\frac{3}{2} lim_{x \to \infty} \frac{1}{2 x e^{x^{2}}}$

Step 4

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

$\frac{3}{2} \cdot \frac{1}{2} lim_{x \to \infty} \frac{1}{x e^{x^{2}}}$

Step 5

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x e^{x^{2}}}$ approaches $0$ .

$\frac{3}{2} \cdot \frac{1}{2} \cdot 0$

Step 6

Simplify the answer.

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Step 6.1

Multiply $\frac{3}{2} \cdot \frac{1}{2}$ .

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Step 6.1.1

Multiply $\frac{3}{2}$ by $\frac{1}{2}$ .

$\frac{3}{2 \cdot 2} \cdot 0$

Step 6.1.2

Multiply $2$ by $2$ .

$\frac{3}{4} \cdot 0$

Step 6.2

Multiply $\frac{3}{4}$ by $0$ .

$0$