What is $$\lim_{x\to 2}\frac{-1}{x^2-4x+4} ?$$
For $lim_{x\to 2^+}$ I get:
$$\frac{-1}{4-8+4} = \frac{-1}{0^+} = -\infty$$
But for $lim_{x\to 2^-}$ I get:
$$\frac{-1}{4-8+4} = \frac{-1}{0^-} = \infty$$
According to this, there should be no limit. However the final answer in my book is $-\infty$
Where was I wrong?
Hint: Notice $$\lim_{x\to 2}\frac{-1}{x^2-4x+4}$$ $$=\lim_{x\to 2}\frac{-1}{(x-2)^2}$$
You may check the limit is $$\lim_{x\to 2^{+}}\frac{-1}{(x-2)^2}=\lim_{x\to 2^{-}}\frac{-1}{(x-2)^2}=-\infty$$
Double check your claim that $$ \frac{-1}{0^-} \equiv \infty$$ I think you may be cancelling the "$-$" superscript on $0$ with the negative one in the numerator. Since you can rewrite $$x^2-4x+4 = (x-2)^2$$ you should never have a negative quantity in the denominator, which will force the overall quantity to be strictly negative.
In the second case, the denominator is also $0^+$. You can check this by calculating $1.99^2-4\times 1.99+4$.
Apart from writing $x^2-4x+4=(x-2)^2$ as suggested by others, you can also try substituting $x=2-\epsilon$ where $\epsilon>0$ and find the limit as $\epsilon \downarrow 0$ (this is for more general cases, say for higher order polynomials when you won't want to factorise). In this case you would end up with $\frac{-1}{\epsilon^2}$; long-winded, but this is a more general approach if you don't like factorising.
