Calculus Examples

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

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.

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

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 $g (x) = 2 x$ .

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Step 1.3.3.1

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

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

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 $a$ = $e$ .

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

Step 1.3.3.3

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

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

Step 1.3.4

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 1.3.5

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

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

Step 1.3.6

Multiply $2$ by $1$ .

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

Step 1.3.7

Move $2$ to the left of $e^{2 x}$ .

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

Step 1.4

Cancel the common factor of $2$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 2

Apply L'Hospital's rule.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 2.2

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

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

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

Step 2.3.2

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

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

Step 2.3.3

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 $g (x) = 2 x$ .

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Step 2.3.3.1

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

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

Step 2.3.3.2

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

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

Step 2.3.3.3

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

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

Step 2.3.4

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 2.3.5

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

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

Step 2.3.6

Multiply $2$ by $1$ .

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

Step 2.3.7

Move $2$ to the left of $e^{2 x}$ .

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

Step 3

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

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

Step 4

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

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

Step 5

Multiply $\frac{1}{2}$ by $0$ .

$0$