Find $\lim_{x \to 1}\frac{\sqrt{x^2+35}-6}{x-1}$

Question

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.

Answer 1

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.)

Answer 2

You may check about l'hospitals rule. If limit of both numerator and denominator tend to zero, you differentiate the numerator and denominator with respect to the variable to which you apply limit.

By doing that you 1/6 I guess.

Check the following link

http://mathworld.wolfram.com/LHospitalsRule.html

Answer 3

You already have been given good answers to the problem.

Let me show you another approach which can be of interest (even just for your curiosity at the present time).

To make life simpler, set $x=1+y$. This makes

$$A=\frac{\sqrt{x^2+35}-6}{x-1}=\frac{\sqrt{y^2+2 y+36}-6}{y}$$ Consider the first portion of the numerator $$\sqrt{y^2+2 y+36}=6 \sqrt{1+\frac y{18}+\frac {y^2}{36}}$$ Since $y\to 0$, term $y^2$ is very negligible and then $$\sqrt{1+\frac y{18}+\frac {y^2}{36}}\approx \sqrt{1+\frac y{18}}$$ May be, you already know that for small values of $z$, $\sqrt{1+z}\approx 1+\frac z2$ (you can easily check it with a pocket calculator). So $$\sqrt{1+\frac y{18}+\frac {y^2}{36}}\approx \sqrt{1+\frac y{18}}\approx1+\frac y{36}$$ All of that makes $$A\approx\frac{6(1+\frac y{36})-6} y=\frac{6+\frac y{6}-6}y=\frac 16$$ which is the limit.