Simplify limit as x approaches 3 of ( square root of x+1-2)/(x-3)

$lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 1

Apply L\'Hospital\'s rule.

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

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

$\frac{lim_{x \to 3} \sqrt{x + 1} - 2}{lim_{x \to 3} x - 3}$

Evaluate the limit of the numerator.

Evaluate the limit.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $3$ .

$\frac{lim_{x \to 3} \sqrt{x + 1} - lim_{x \to 3} 2}{lim_{x \to 3} x - 3}$

Move the limit under the radical sign.

$\frac{\sqrt{lim_{x \to 3} x + 1} - lim_{x \to 3} 2}{lim_{x \to 3} x - 3}$

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $3$ .

$\frac{\sqrt{lim_{x \to 3} x + lim_{x \to 3} 1} - lim_{x \to 3} 2}{lim_{x \to 3} x - 3}$

Evaluate the limit of $1$ which is constant as $x$ approaches $3$ .

$\frac{\sqrt{lim_{x \to 3} x + 1} - lim_{x \to 3} 2}{lim_{x \to 3} x - 3}$

Evaluate the limit of $2$ which is constant as $x$ approaches $3$ .

$\frac{\sqrt{lim_{x \to 3} x + 1} - 1 \cdot 2}{lim_{x \to 3} x - 3}$

Evaluate the limit of $x$ by plugging in $3$ for $x$ .

$\frac{\sqrt{3 + 1} - 1 \cdot 2}{lim_{x \to 3} x - 3}$

Simplify the answer.

Simplify each term.

Add $3$ and $1$ .

$\frac{\sqrt{4} - 1 \cdot 2}{lim_{x \to 3} x - 3}$

Rewrite $4$ as $2^{2}$ .

$\frac{\sqrt{2^{2}} - 1 \cdot 2}{lim_{x \to 3} x - 3}$

Pull terms out from under the radical, assuming positive real numbers.

$\frac{2 - 1 \cdot 2}{lim_{x \to 3} x - 3}$

Multiply $- 1$ by $2$ .

$\frac{2 - 2}{lim_{x \to 3} x - 3}$

Subtract $2$ from $2$ .

$\frac{0}{lim_{x \to 3} x - 3}$

Evaluate the limit of the denominator.

Evaluate the limit.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $3$ .

$\frac{0}{lim_{x \to 3} x - lim_{x \to 3} 3}$

Evaluate the limit of $3$ which is constant as $x$ approaches $3$ .

$\frac{0}{lim_{x \to 3} x - 1 \cdot 3}$

Evaluate the limit of $x$ by plugging in $3$ for $x$ .

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

Simplify the answer.

Multiply $- 1$ by $3$ .

$\frac{0}{3 - 3}$

Subtract $3$ from $3$ .

$\frac{0}{0}$

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Since $\frac{0}{0}$ 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.

$lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = lim_{x \to 3} \frac{\frac{d}{d x} [\sqrt{x + 1} - 2]}{\frac{d}{d x} [x - 3]}$

Find the derivative of the numerator and denominator.

Differentiate the numerator and denominator.

$lim_{x \to 3} \frac{\frac{d}{d x} [\sqrt{x + 1} - 2]}{\frac{d}{d x} [x - 3]}$

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

$lim_{x \to 3} \frac{\frac{d}{d x} [\sqrt{x + 1}] + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

Evaluate $\frac{d}{d x} [\sqrt{x + 1}]$ .

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

$lim_{x \to 3} \frac{\frac{d}{d x} [{(x + 1)}^{\frac{1}{2}}] + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

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

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

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

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

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

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

$lim_{x \to 3} \frac{\frac{1}{2} {(x + 1)}^{\frac{1}{2} - 1} \frac{d}{d x} [x + 1] + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

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

$lim_{x \to 3} \frac{\frac{1}{2} {(x + 1)}^{\frac{1}{2} - 1} (\frac{d}{d x} [x] + \frac{d}{d x} [1]) + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

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

$lim_{x \to 3} \frac{\frac{1}{2} {(x + 1)}^{\frac{1}{2} - 1} (1 + \frac{d}{d x} [1]) + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

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

$lim_{x \to 3} \frac{\frac{1}{2} {(x + 1)}^{\frac{1}{2} - 1} (1 + 0) + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

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

$lim_{x \to 3} \frac{\frac{1}{2} {(x + 1)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (1 + 0) + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

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

$lim_{x \to 3} \frac{\frac{1}{2} {(x + 1)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (1 + 0) + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

Combine the numerators over the common denominator.

$lim_{x \to 3} \frac{\frac{1}{2} {(x + 1)}^{\frac{1 - 1 \cdot 2}{2}} (1 + 0) + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$lim_{x \to 3} \frac{\frac{1}{2} {(x + 1)}^{\frac{1 - 2}{2}} (1 + 0) + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

Subtract $2$ from $1$ .

$lim_{x \to 3} \frac{\frac{1}{2} {(x + 1)}^{\frac{- 1}{2}} (1 + 0) + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

Move the negative in front of the fraction.

$lim_{x \to 3} \frac{\frac{1}{2} {(x + 1)}^{- \frac{1}{2}} (1 + 0) + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

Add $1$ and $0$ .

$lim_{x \to 3} \frac{\frac{1}{2} {(x + 1)}^{- \frac{1}{2}} \cdot 1 + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

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

$lim_{x \to 3} \frac{\frac{{(x + 1)}^{- \frac{1}{2}}}{2} \cdot 1 + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

Multiply $\frac{{(x + 1)}^{- \frac{1}{2}}}{2}$ by $1$ .

$lim_{x \to 3} \frac{\frac{{(x + 1)}^{- \frac{1}{2}}}{2} + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

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

$lim_{x \to 3} \frac{\frac{1}{2 {(x + 1)}^{\frac{1}{2}}} + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 3]}$

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$lim_{x \to 3} \frac{\frac{1}{2 {(x + 1)}^{\frac{1}{2}}} + 0}{\frac{d}{d x} [x - 3]}$

Add $\frac{1}{2 {(x + 1)}^{\frac{1}{2}}}$ and $0$ .

$lim_{x \to 3} \frac{\frac{1}{2 {(x + 1)}^{\frac{1}{2}}}}{\frac{d}{d x} [x - 3]}$

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$lim_{x \to 3} \frac{\frac{1}{2 {(x + 1)}^{\frac{1}{2}}}}{\frac{d}{d x} [x] + \frac{d}{d x} [- 3]}$

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

$lim_{x \to 3} \frac{\frac{1}{2 {(x + 1)}^{\frac{1}{2}}}}{1 + \frac{d}{d x} [- 3]}$

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$lim_{x \to 3} \frac{\frac{1}{2 {(x + 1)}^{\frac{1}{2}}}}{1 + 0}$

Add $1$ and $0$ .

$lim_{x \to 3} \frac{\frac{1}{2 {(x + 1)}^{\frac{1}{2}}}}{1}$

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 3} \frac{1}{2 {(x + 1)}^{\frac{1}{2}}} \cdot 1$

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

$lim_{x \to 3} \frac{1}{2 \sqrt{x + 1}} \cdot 1$

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

$lim_{x \to 3} \frac{1}{2 \sqrt{x + 1}}$

Step 2

Evaluate the limit.

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

$\frac{1}{2} lim_{x \to 3} \frac{1}{\sqrt{x + 1}}$

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $3$ .

$\frac{1}{2} \cdot \frac{lim_{x \to 3} 1}{lim_{x \to 3} \sqrt{x + 1}}$

Evaluate the limit of $1$ which is constant as $x$ approaches $3$ .

$\frac{1}{2} \cdot \frac{1}{lim_{x \to 3} \sqrt{x + 1}}$

Move the limit under the radical sign.

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to 3} x + 1}}$

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $3$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to 3} x + lim_{x \to 3} 1}}$

Evaluate the limit of $1$ which is constant as $x$ approaches $3$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{lim_{x \to 3} x + 1}}$

Step 3

Simplify the denominator.

Evaluate the limit of $x$ by plugging in $3$ for $x$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{3 + 1}}$

Add $3$ and $1$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{4}}$

Rewrite $4$ as $2^{2}$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{2^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

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

Step 4

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

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

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

Multiply $2$ by $2$ .

$\frac{1}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{4}$

Decimal Form:

$0.25$

Simplify limit as x approaches 3 of ( square root of x+1-2)/(x-3)

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