Evaluate Using L’Hospital’s Rule limit as x approaches infinity of x/( square root of x^2+1)

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

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

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

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

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

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

As $x$ approaches $\infty$ for radicals, the value goes to $\infty$ .

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

Infinity divided by infinity is undefined.

Undefined

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

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.

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

Find the derivative of the numerator and denominator.

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Differentiate the numerator and denominator.

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

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

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

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + 1}$ as ${(x^{2} + 1)}^{\frac{1}{2}}$ .

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

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

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To apply the Chain Rule, set $u$ as $x^{2} + 1$ .

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

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

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

Replace all occurrences of $u$ with $x^{2} + 1$ .

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Combine $- 1$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Move the negative in front of the fraction.

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

Combine $\frac{1}{2}$ and ${(x^{2} + 1)}^{- \frac{1}{2}}$ .

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

Move ${(x^{2} + 1)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

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

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

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

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

$lim_{x \to \infty} \frac{1}{\frac{1}{2 {(x^{2} + 1)}^{\frac{1}{2}}} (2 x + 0)}$

Add $2 x$ and $0$ .

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

Combine $2$ and $\frac{1}{2 {(x^{2} + 1)}^{\frac{1}{2}}}$ .

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

Combine $\frac{2}{2 {(x^{2} + 1)}^{\frac{1}{2}}}$ and $x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply the numerator by the reciprocal of the denominator.

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

Rewrite ${(x^{2} + 1)}^{\frac{1}{2}}$ as $\sqrt{x^{2} + 1}$ .

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

Multiply $\frac{\sqrt{x^{2} + 1}}{x}$ by $1$ .

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

Take the limit of each term.

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Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $\sqrt{x^{2}} = x$ .

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

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

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Cancel the common factor.

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

Divide $1$ by $1$ .

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

Cancel the common factor of $x$ .

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Cancel the common factor.

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

Divide $1$ by $1$ .

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

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

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

Move the limit under the radical sign.

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

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

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

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

$\frac{\sqrt{lim_{x \to \infty} 1 + 0}}{lim_{x \to \infty} 1}$

Evaluate the limits by plugging in the value for the variable.

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

$\frac{\sqrt{1 + 0}}{lim_{x \to \infty} 1}$

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

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

Simplify the answer.

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Divide $\sqrt{1 + 0}$ by $1$ .

$\sqrt{1 + 0}$

Add $1$ and $0$ .

$\sqrt{1}$

Any root of $1$ is $1$ .

$1$

Evaluate Using L’Hospital’s Rule limit as x approaches infinity of x/( square root of x^2+1)

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