Math / Calculus

QuestionEvaluate $\lim _{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x^{2}-49}$

Studdy Solution

STEP 1

1. We are given the limit expression $\lim _{x \rightarrow 7} \frac{\sqrt{x+2}-3}{x^{2}-49}$ .
2. We need to evaluate this limit as $x$ approaches 7.

STEP 2

1. Substitute $x = 7$ into the expression to check for indeterminate form.
2. Simplify the expression using algebraic techniques.
3. Re-evaluate the limit after simplification.

STEP 3

Substitute $x = 7$ into the expression:
$\frac{\sqrt{7+2}-3}{7^{2}-49} = \frac{\sqrt{9}-3}{49-49} = \frac{3-3}{0} = \frac{0}{0}$
The expression is in the indeterminate form $\frac{0}{0}$ .

STEP 4

Simplify the expression by rationalizing the numerator. Multiply the numerator and the denominator by the conjugate of the numerator:
$\frac{\sqrt{x+2}-3}{x^{2}-49} \cdot \frac{\sqrt{x+2}+3}{\sqrt{x+2}+3} = \frac{(\sqrt{x+2}-3)(\sqrt{x+2}+3)}{(x^{2}-49)(\sqrt{x+2}+3)}$

STEP 5

Simplify the numerator using the difference of squares:
$(\sqrt{x+2}-3)(\sqrt{x+2}+3) = (\sqrt{x+2})^2 - 3^2 = x+2 - 9 = x - 7$

STEP 6

Factor the denominator:
$x^{2} - 49 = (x-7)(x+7)$

STEP 7

Substitute the simplified numerator and denominator back into the expression:
$\frac{x - 7}{(x-7)(x+7)(\sqrt{x+2}+3)}$
Cancel the common factor $(x-7)$ :
$\frac{1}{(x+7)(\sqrt{x+2}+3)}$

STEP 8

Re-evaluate the limit as $x$ approaches 7:
$\lim _{x \rightarrow 7} \frac{1}{(x+7)(\sqrt{x+2}+3)} = \frac{1}{(7+7)(\sqrt{7+2}+3)}$
Simplify:
$= \frac{1}{14 \times (3+3)} = \frac{1}{14 \times 6} = \frac{1}{84}$
The value of the limit is:
$\boxed{\frac{1}{84}}$

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