Question

chapter 1 limits and their properties
60

1. $lim_{x

ightarrow2}\frac{|x - 2|}{x - 2}$

1. $lim_{x

ightarrow5}\frac{2}{x - 5}$

1. $lim_{x

ightarrow0}cos\frac{1}{x}$

1. $lim_{x

ightarrowpi/2}\tan x$

Explanation:

Step1: Analyze left - hand limit for $\lim_{x

ightarrow2}\frac{|x - 2|}{x - 2}$
When $x
ightarrow2^{-}$, $x-2<0$, so $|x - 2|=-(x - 2)$. Then $\lim_{x
ightarrow2^{-}}\frac{|x - 2|}{x - 2}=\lim_{x
ightarrow2^{-}}\frac{-(x - 2)}{x - 2}=- 1$.

Step2: Analyze right - hand limit for $\lim_{x

ightarrow2}\frac{|x - 2|}{x - 2}$
When $x
ightarrow2^{+}$, $x - 2>0$, so $|x - 2|=x - 2$. Then $\lim_{x
ightarrow2^{+}}\frac{|x - 2|}{x - 2}=\lim_{x
ightarrow2^{+}}\frac{x - 2}{x - 2}=1$.
Since the left - hand limit and the right - hand limit are not equal, $\lim_{x
ightarrow2}\frac{|x - 2|}{x - 2}$ does not exist.

Step3: Analyze $\lim_{x

ightarrow5}\frac{2}{x - 5}$
As $x
ightarrow5^{+}$, $x-5
ightarrow0^{+}$, then $\lim_{x
ightarrow5^{+}}\frac{2}{x - 5}=+\infty$. As $x
ightarrow5^{-}$, $x - 5
ightarrow0^{-}$, then $\lim_{x
ightarrow5^{-}}\frac{2}{x - 5}=-\infty$. So $\lim_{x
ightarrow5}\frac{2}{x - 5}$ does not exist.

Step4: Analyze $\lim_{x

ightarrow0}\cos\frac{1}{x}$
Let $t=\frac{1}{x}$. As $x
ightarrow0$, $t
ightarrow\pm\infty$. The function $y = \cos t$ oscillates between $- 1$ and $1$ as $t
ightarrow\pm\infty$. So $\lim_{x
ightarrow0}\cos\frac{1}{x}$ does not exist.

Step5: Analyze $\lim_{x

ightarrow\frac{\pi}{2}}\tan x$
We know that $\tan x=\frac{\sin x}{\cos x}$. As $x
ightarrow\frac{\pi}{2}^{+}$, $\cos x
ightarrow0^{-}$ and $\sin x
ightarrow1$, so $\lim_{x
ightarrow\frac{\pi}{2}^{+}}\tan x=-\infty$. As $x
ightarrow\frac{\pi}{2}^{-}$, $\cos x
ightarrow0^{+}$ and $\sin x
ightarrow1$, so $\lim_{x
ightarrow\frac{\pi}{2}^{-}}\tan x=+\infty$. So $\lim_{x
ightarrow\frac{\pi}{2}}\tan x$ does not exist.

Answer:

1. Does not exist
2. Does not exist
3. Does not exist
4. Does not exist