Question:
Evaluate the Given limit: \(\lim_{x\rightarrow -2}\) \(\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\)
Evaluate the Given limit: \(\lim_{x\rightarrow -2}\) \(\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\)
Updated On: Oct 23, 2023
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Solution and Explanation
At x = -2, the value of the given function takes the form 0/0.
\(\lim_{x\rightarrow -2}\) \(\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\)=\(\lim_{x\rightarrow -2}\) \(\frac{\frac{2+x}{2x}}{x+2}\)
= \(\lim_{x\rightarrow -2}\) \(\frac{1}{2x}\)
= \(\frac{1}{2(-2)}\)
= \(-\frac{1}{4}\)
\(\lim_{x\rightarrow -2}\) \(\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\)=\(\lim_{x\rightarrow -2}\) \(\frac{\frac{2+x}{2x}}{x+2}\)
= \(\lim_{x\rightarrow -2}\) \(\frac{1}{2x}\)
= \(\frac{1}{2(-2)}\)
= \(-\frac{1}{4}\)
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