Calculus Examples

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

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.

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $2$ .

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

Step 2

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

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

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.

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

Step 3.1.2

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

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

Step 3.1.3

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

$\frac{2}{\ln (2)} \cdot \frac{\infty}{\infty}$

Step 3.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{2}{\ln (2)} \cdot \frac{\infty}{\infty}$

Step 3.2

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

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

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

Step 3.3.2

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

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

Step 3.3.3

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $2$ .

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

Step 4

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

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

Step 5

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

$\frac{2}{\ln (2)} \cdot \frac{1}{\ln (2)} \cdot 0$

Step 6

Simplify the answer.

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Step 6.1

Multiply $\frac{2}{\ln (2)} \cdot \frac{1}{\ln (2)}$ .

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Step 6.1.1

Multiply $\frac{2}{\ln (2)}$ by $\frac{1}{\ln (2)}$ .

$\frac{2}{\ln (2) \ln (2)} \cdot 0$

Step 6.1.2

Raise $\ln (2)$ to the power of $1$ .

$\frac{2}{\ln^{1} (2) \ln (2)} \cdot 0$

Step 6.1.3

Raise $\ln (2)$ to the power of $1$ .

$\frac{2}{\ln^{1} (2) \ln^{1} (2)} \cdot 0$

Step 6.1.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{2}{{\ln (2)}^{1 + 1}} \cdot 0$

Step 6.1.5

Add $1$ and $1$ .

$\frac{2}{\ln^{2} (2)} \cdot 0$

Step 6.2

Multiply $\frac{2}{\ln^{2} (2)}$ by $0$ .

$0$