# Question #36d3b

Apr 6, 2017

#### Explanation:

As typed, the question asks for ${\lim}_{x \rightarrow 1} {\left(x - 1\right)}^{2} \cos \left(\frac{1}{x} - 1\right)$.

Interpretation 1
If you really want ${\lim}_{x \rightarrow 1} {\left(x - 1\right)}^{2} \cos \left(\frac{1}{x} - 1\right)$, use substitution to get

${\left(1 - 1\right)}^{2} \cos \left(\frac{1}{1} - 1\right) = 0 \cos \left(0\right) = 0 \cdot 1 = 0$

Interpretation 2

For ${\lim}_{x \rightarrow 1} {\left(x - 1\right)}^{2} \cos \left(\frac{1}{x - 1}\right)$ argue something like this:

We know from our study of trigonometry, that

$- 1 \le \cos \left(\frac{1}{x - 1}\right) \le 1$ for all $x \ne 1$.

Since ${\left(x - 1\right)}^{2}$ is positive for all $x \ne 1$, we can multiply the inequality without changing the direction of the inequality.

$- {\left(x - 1\right)}^{2} \le {\left(x - 1\right)}^{2} \cos \left(\frac{1}{x - 1}\right) \le {\left(x - 1\right)}^{2}$ for $x \ne 1$

Note that

${\lim}_{x \rightarrow 1} \left(- {\left(x - 1\right)}^{2}\right) = 0$ and ${\lim}_{x \rightarrow 1} {\left(x - 1\right)}^{2} = 0$.

Therefore, by the squeeze theorem,

${\lim}_{x \rightarrow 1} {\left(x - 1\right)}^{2} \cos \left(\frac{1}{x - 1}\right)$