Have some trouble with limits

Question

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

Why would it not exist when $x\to 2^-$? Also I don't get your $2)$, $x\to 1$, how could $x = 0$ or $-1$? — user88595, Apr 03 '14 at 20:17
it surely can but you get a change of sign. Think through it carefully. — user88595, Apr 03 '14 at 20:22
Do you know about Taylor series? That would help a lot with (3) and (4). Factor the denominator of (2) at $a^3-b^3$ with $a = \sqrt[3]{x}, b= 1$. — gt6989b, Apr 03 '14 at 20:29
if it can cancle out, doest it mean the limit is 1 and that's the answer? — user136877, Apr 03 '14 at 20:29
I don't think I have learnt Taylor series. Is there an alternative way of doing these? — user136877, Apr 03 '14 at 20:35

score 0 · Accepted Answer · answered Apr 03 '14 at 21:17

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

For #4, I believe you mean that the second factor oscillates between $ \ -x \ $ and $ \ +x \ $ , the limit being zero by the "Squeeze Theorem" (or "Sandwich Theorem" or "Pinching Theorem" or whatever it's called where you are). — colormegone, Apr 03 '14 at 21:38
You're right, I'll edit my answer! (We call it either the 2 policemen theorem or the comparison theorem) — Alessandro Codenotti, Apr 03 '14 at 21:54
how did this $\frac{x^{100}+1}{\sqrt{x^{200}+x^{100}+1}+x^{100}}$ become $\frac{x^{100}}{2x^{100}}$ — user136877, Apr 03 '14 at 22:13
$\sqrt{x^{200}+x^{100}+1}+x^{100}=\sqrt{x^{200}}+x^{100}=x^{100}+x^{100}$ (we can do this steps because we're taking a limit to infinity). The limit as $x\to 0$ of both $\frac{sin(x)}{x}$ and $\frac{x}{sin(x)}$ is $1$. — Alessandro Codenotti, Apr 04 '14 at 05:29

Have some trouble with limits

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