Compute a limit or prove that it does not exist

Question

Do the following limits exist? Compute them or prove that they do not exist.

(a) $\lim_{x\to 1}\frac{x^2-x}{2x^2-x-1}$
(b) $\lim_{x\to 1}\frac{|x-1|}{2x^2-x-1}$

For (a) it's pretty easy to see that the limit exist and it is $\frac13$, it's just $\frac{x(x-1)}{(2x +1)(x-1)} =\frac{ x }{ 2x +1 }= \frac{1}{3}$.

I'm guessing that (b) has no limit, but I can't find a way to prove it.

Answer 1

For (b) :

Note that we have $|x-1|=x-1$ for $x\ge 1$, and $|x-1|=-(x-1)$ for $x\lt 1$.

So, you have $$\lim_{x\to 1^+}\frac{|x-1|}{2x^2-x-1}=\lim_{x\to 1^+}\frac{x-1}{(2x+1)(x-1)}$$ and $$\lim_{x\to 1^-}\frac{|x-1|}{2x^2-x-1}=\lim_{x\to 1^-}\frac{-(x-1)}{(2x+1)(x-1)}$$

Answer 2

HINTS:

a)

$$\lim_{x\to 1}\frac{x^2}{2x^2-x-1}=\lim_{x\to 1}\frac{(x-1)x}{x-1)(2x+1)}=\lim_{x\to 1}\frac{x}{2x+1}$$

b)

$$\lim_{x\to 1^+}\frac{|x-1|}{2x^2-x-1}=\lim_{x\to 1^+}\frac{\frac{1}{2x+1}|x-1|}{x-1}=\left(\lim_{x\to 1^+}\frac{1}{2x+1}\right)\left(\lim_{x\to 1^+}\frac{|x-1|}{x-1}\right)=$$ $$\left(\frac{1}{2\cdot 1+1}\right)\left(\lim_{x\to 1^+}\frac{|x-1|}{x-1}\right)=\left(\frac{1}{3}\right)\left(\lim_{x\to 1^+}\frac{|x-1|}{x-1}\right)$$

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Let $u=x-1$. Then $\frac{|x-1|}{x-1}=\lim_{u\to 0^+}\frac{|u|}{u}$.

So, since $u\to 0$ from the right as $x\to 1$ from the right, we have that:

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$$\frac{1}{3}\lim_{x\to 1^+}\frac{|x-1|}{x-1}=\frac{\lim_{u\to 0^+}\frac{|u|}{u}}{3}=\frac{\lim_{u\to 0^+}\frac{\frac{\text{d}}{\text{d}u}|u|}{\frac{\text{d}}{\text{d}u}u}}{3}=\frac{\lim_{u\to 0^+}\frac{\text{d}}{\text{d}u}|u|}{3}$$