How To Solve A Limit Using The Squeeze Theorem

lim x->1 (x-1)^2 cos (1/x-1)

Calculus
Helpmewithmath Helpmewithmath
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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
\[\lim_{x\rightarrow 0} (x-1)^2\cos (\frac{1}{x}-1)=0.\]

Erdos Erdos
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