Limit of a natural logarithm

Question

Can someone give me some guidelines as to how to find the following limit? $$\lim_{h\to 2}\ln(h^2-2h)$$ I know if $h=2$ had to be plugged in, it would give $ln(0)$ which does not exist, but when I used an online limit calculator it says the limit is at $-∞$? I need some help with the steps to finding $-∞$.

Answer 1

Note that

• for $h\to2^+$

$$\ln(h^2-2h)=\ln h(h-2)=\ln h + \ln(h-2)\to\ln 2-\infty=-\infty $$

• for $h\to2^-$

the expression in not well defined.

Answer 2

Hint

Make $h=x+2$ to get $$h^2-2h=x(x+2)$$ Now, what happens when $x\to 0$ ?