Have some trouble with limits Find the following limits， do not use L’Hopital’s Rule. If the limit does not exist, explain
why.


*

*$$\lim_{x\to 2}\frac{\left | x-2 \right |}{x^2-3x+2}$$
solution: $\lim_{x\to 2}\frac{\left | x-2 \right |}{(x-1)(x-2)}$,does this mean that when $x\rightarrow 2^+$, limit exist,which is 1. when  $x\rightarrow 2^-$ ,limit doesn't exist?

*$$\lim_{x\to 1}\frac{\sqrt[3]{x} - 1}{x-1}$$
solution: The limit only exist when $x=0,x=-1$?

*$$\lim_{x\to \infty }(\sqrt{x^{200}+x^{100}+1}）-x^{100}$$
solution: Don't know how to do this one,find derivative?

*$$\lim_{x\to 0}\frac{x^2 \sin 1/x}{\sin x}$$
solution: since $-1 \le \sin x \le 1$,$-x^2 \le x^2 \sin x \le x^2$
just want to know If my answer is right
 A: 1 as you already solved$\lim\limits_{x\to2^+}\frac{|x-2|}{(x-1)(x-2)}=1$
when you do the limit as $x\to2^-$ it is the same, but you have to pay attentions to the signs, you  can cancel out the $x-2$ but you need to put a minus in front of the fraction, the limit is then $-1$.
2 if we substitute $x=1$ we obtain $\frac{0}{0}$ which is an indeterminate form, to get rid of it we need to rationalize the fraction, since we have a cube root at the numerator we will use the identity $a^3-b^3=(a-b)(a^2+b^2+ab)$:
$\frac{\sqrt[3]{x}-1}{x-1}\cdot\frac{\sqrt[3]{x^2}+1+\sqrt[3]{x}}{\sqrt[3]{x^2}+1+\sqrt[3]{x}}=\frac{x-1}{(x-1)(\sqrt[3]{x^2}+1+\sqrt[3]{x})}$
now you can cancel out the $x-1$ and then take the limit, which should be $\frac{1}{3}$
3 When you are taking a limit to infity of a root the only term that matter is the one with the highest degree because it'll be much bigger compared to the others.

$\lim\limits_{x\to\infty}\sqrt{x^{200}+x^{100}+1}-x^{100}=\lim\limits_{x\to\infty}\sqrt{x^{200}}-x^{100}=\lim\limits_{x\to\infty}x^{100}-x^{100}=\infty-\infty$
This is another indeterminate form and, once again, we need to rationalize:

$\sqrt{x^{200}+x^{100}+1}-x^{100}\cdot\frac{\sqrt{x^{200}+x^{100}+1}+x^{100}}{\sqrt{x^{200}+x^{100}+1}+x^{100}}=\frac{x^{100}+1}{\sqrt{x^{200}+x^{100}+1}+x^{100}}$
as I said we need only the terms of highest degree so we only have to take the limits of $\frac{x^{100}}{2x^{100}}$ which is $\frac{1}{2}$
4 this one is easier, just rewrite it as $\frac{x}{sin(x)}\cdot(xsin(\frac{1}{x}))$, the left term is a very known limit, and its value is 1, the right term is zero times sin(1/x) which is oscilatting between -1 and 1 so the limit is 0
