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

4.8K
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- 1 Answer
- 505 views
- Pro Bono
Related Questions
- Integrate $\int x^2(1-x^2)^{-\frac{3}{2}}dx$
- Show that $\sum_{n=1}^{\infty} \frac{\sin n}{n}$ is convergent
- (Calculus 1) Basic Calc: Derivatives, optimization, linear approximation...
- Prove that ${n\choose 2}2^{n-2}=\sum\limits_{k=2}^{n}{n\choose k}{k\choose 2}$ for all $n\geq 2$
- taking business calc and prin of finance class should i buy calculator in body
- Calculus problems on improper integrals
- Minimizing the cost of building a box
- Extremal values/asymptotes