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

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

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.

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

Evaluate the limit of the numerator.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

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

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

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

Simplify the expression.

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

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

Subtract $2$ from $2$ .

$\frac{0}{lim_{x \to 2} \sqrt{x + 2} - 2}$

Evaluate the limit of the denominator.

Evaluate the limit.

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{0}{lim_{x \to 2} \sqrt{x + 2} - lim_{x \to 2} 2}$

Move the limit under the radical sign.

$\frac{0}{\sqrt{lim_{x \to 2} x + 2} - lim_{x \to 2} 2}$

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{0}{\sqrt{lim_{x \to 2} x + lim_{x \to 2} 2} - lim_{x \to 2} 2}$

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

$\frac{0}{\sqrt{lim_{x \to 2} x + 2} - lim_{x \to 2} 2}$

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

$\frac{0}{\sqrt{lim_{x \to 2} x + 2} - 1 \cdot 2}$

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{0}{\sqrt{2 + 2} - 1 \cdot 2}$

Simplify the answer.

Simplify each term.

Add $2$ and $2$ .

$\frac{0}{\sqrt{4} - 1 \cdot 2}$

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

$\frac{0}{\sqrt{2^{2}} - 1 \cdot 2}$

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

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

Multiply $- 1$ by $2$ .

$\frac{0}{2 - 2}$

Subtract $2$ from $2$ .

$\frac{0}{0}$

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.

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

Find the derivative of the numerator and denominator.

Differentiate the numerator and denominator.

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

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

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

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

$lim_{x \to 2} \frac{0 + \frac{d}{d x} [- x]}{\frac{d}{d x} [\sqrt{x + 2} - 2]}$

Evaluate $\frac{d}{d x} [- x]$ .

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

$lim_{x \to 2} \frac{0 - \frac{d}{d x} [x]}{\frac{d}{d x} [\sqrt{x + 2} - 2]}$

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

$lim_{x \to 2} \frac{0 - 1 \cdot 1}{\frac{d}{d x} [\sqrt{x + 2} - 2]}$

Multiply $- 1$ by $1$ .

$lim_{x \to 2} \frac{0 - 1}{\frac{d}{d x} [\sqrt{x + 2} - 2]}$

Subtract $1$ from $0$ .

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

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

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

Evaluate $\frac{d}{d x} [\sqrt{x + 2}]$ .

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

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

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 $g (x) = x + 2$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Move the negative in front of the fraction.

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

Add $1$ and $0$ .

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

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

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

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

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

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

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

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

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

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

$lim_{x \to 2} - (2 \sqrt{x + 2})$

Multiply $2$ by $- 1$ .

$lim_{x \to 2} - 2 \sqrt{x + 2}$

Step 2

Evaluate the limit.

Move the term $- 2$ outside of the limit because it is constant with respect to $x$ .

$- 2 lim_{x \to 2} \sqrt{x + 2}$

Move the limit under the radical sign.

$- 2 \sqrt{lim_{x \to 2} x + 2}$

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$- 2 \sqrt{lim_{x \to 2} x + lim_{x \to 2} 2}$

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

$- 2 \sqrt{lim_{x \to 2} x + 2}$

Step 3

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$- 2 \sqrt{2 + 2}$

Step 4

Simplify the expression.

Add $2$ and $2$ .

$- 2 \sqrt{4}$

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

$- 2 \sqrt{2^{2}}$

Step 5

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

$- 2 \cdot 2$

Step 6

Multiply $- 2$ by $2$ .

$- 4$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.