Question

Find the limits: lim_{xrightarrow2}frac{sqrt{x^2+12}-4}{x-2}

Limits and continuity
ANSWERED
asked 2020-11-10
Find the limits:
\(\lim_{x\rightarrow2}\frac{\sqrt{x^2+12}-4}{x-2}\)

Answers (1)

2020-11-11

We have to find limits:
\(\lim_{x\rightarrow2}\frac{\sqrt{x^2+12}-4}{x-2}\)
Since we have an indeterminate form of type \(\frac{0}{0}\), we can apply the L'hospital rule:
\(\lim_{x\rightarrow2}\frac{\sqrt{x^2+12}-4}{x-2}=\lim_{x\rightarrow2}\frac{\frac{d}{dx}(\sqrt{x^2+12}-4)}{\frac{d}{dx}(x-2)}\)
\(\lim_{x\rightarrow2}\frac{\frac{2x}{2\sqrt{x^2+12}}}{1}\)
\(\lim_{x\rightarrow2}\frac{x}{\sqrt{x^2+12}}=\frac{2}{\sqrt{2^2+12}}=\frac{2}{\sqrt{4+12}}\)
\(=\frac{2}{\sqrt{16}}=\frac{2}{4}=\frac{1}{2}\)
\(\Rightarrow\lim_{x\rightarrow2}\frac{\sqrt{x^2+12}-4}{x-2}=\frac{1}{2}\)

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