Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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

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

Step 1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

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

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

Step 3.3

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

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

Step 4

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

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Step 4.1

Factor $2$ out of $2^{x} \ln (2)$ .

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

Step 4.2

Cancel the common factors.

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Step 4.2.1

Factor $2$ out of $2 x$ .

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

Step 4.2.2

Cancel the common factor.

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

Step 4.2.3

Rewrite the expression.

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

Step 5

Apply L'Hospital's rule.

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Step 5.1

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

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Step 5.1.1

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

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

Step 5.1.2

Since the function $2^{x - 1}$ approaches $\infty$ , the positive constant $\ln (2)$ times the function also approaches $\infty$ .

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Step 5.1.2.1

Consider the limit with the constant multiple $\ln (2)$ removed.

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

Step 5.1.2.2

Since the exponent $x - 1$ approaches $\infty$ , the quantity $2^{x - 1}$ approaches $\infty$ .

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

Step 5.1.3

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

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

Step 5.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 5.2

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

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

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

Step 5.3.2

Since $\ln (2)$ is constant with respect to $x$ , the derivative of $2^{x - 1} \ln (2)$ with respect to $x$ is $\ln (2) \frac{d}{d x} [2^{x - 1}]$ .

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

Step 5.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 2^{x}$ and $g (x) = x - 1$ .

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Step 5.3.3.1

To apply the Chain Rule, set $u$ as $x - 1$ .

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

Step 5.3.3.2

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

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

Step 5.3.3.3

Replace all occurrences of $u$ with $x - 1$ .

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

Step 5.3.4

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

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

Step 5.3.5

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

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

Step 5.3.6

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

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

Step 5.3.7

Add $1$ and $1$ .

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

Step 5.3.8

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

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

Step 5.3.9

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

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

Step 5.3.10

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

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

Step 5.3.11

Add $1$ and $0$ .

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

Step 5.3.12

Multiply $2^{x - 1}$ by $1$ .

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

Step 5.3.13

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

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

Step 5.4

Divide $2^{x - 1} \ln^{2} (2)$ by $1$ .

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

Step 6

Since the function $2^{x - 1}$ approaches $\infty$ , the positive constant $\ln^{2} (2)$ times the function also approaches $\infty$ .

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Step 6.1

Consider the limit with the constant multiple $\ln^{2} (2)$ removed.

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

Step 6.2

Since the exponent $x - 1$ approaches $\infty$ , the quantity $2^{x - 1}$ approaches $\infty$ .

$\infty$