Question
Find the indicated limit.$\lim _{x \rightarrow-1} \frac{\sqrt{x^{2}+8}}{2 x+4}$
Step 1
First, let's simplify the expression inside the square root: $x^2 + 8 = (x^2 + 2x + 1) + 7 = (x+1)^2 + 7$ Now, the expression becomes: $\lim _{x \rightarrow-1} \frac{\sqrt{(x+1)^2 + 7}}{2x+4}$ Show more…
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