Find $\lim_{x \to 1}\frac{\sqrt{x^2+35}-6}{x-1}$ I've rationalized the numerator to $x^2+35-36$ $$= \frac{x^2+35-36}{(x-1)\left(\sqrt{x^2+35}+6\right)} = \frac00$$ (when I substitute $x=1$) I don't know what to do to the denominator so I can substitute $x=1$ to find the limit.

**Hint 1:** $$35 - 36 = -1$$

**Hint 2:**

You know that the numerator is $0$ when you substitute $x=1$. That means that $(x-1)$ is a factor of the numerator:

$$x^2+35-36 = (x-1)(\ \ldots\ )$$

You can then cancel out $(x-1)$ from top and bottom. (But simplify first using Hint 1.)