How To Solve A Limit Using The Squeeze Theorem
1 Answer
Note that $-1 \leq \cos (\frac{1}{x}-1)\leq 1$. Hence
\[-(x-1)^2\leq (x-1)^2\cos (\frac{1}{x}-1) \leq (x-1)^2.\]
Hence
\[0=\lim_{x\rightarrow 0}-(x-1)^2\leq \lim_{x\rightarrow 0} (x-1)^2\cos (\frac{1}{x}-1) \leq \lim_{x\rightarrow 0} (x-1)^2=0.\]
By the Squeeze Theorem
\[\lim_{x\rightarrow 0} (x-1)^2\cos (\frac{1}{x}-1)=0.\]
\[-(x-1)^2\leq (x-1)^2\cos (\frac{1}{x}-1) \leq (x-1)^2.\]
Hence
\[0=\lim_{x\rightarrow 0}-(x-1)^2\leq \lim_{x\rightarrow 0} (x-1)^2\cos (\frac{1}{x}-1) \leq \lim_{x\rightarrow 0} (x-1)^2=0.\]
By the Squeeze Theorem
\[\lim_{x\rightarrow 0} (x-1)^2\cos (\frac{1}{x}-1)=0.\]
Erdos
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