Evaluate the limit: $\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}$

Question

$$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$

I want to know how to evaluate without using L'Hopital Rule. I'm unable to factorise or simplify it suitably.

Answer 1

$$\frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} = 1+ \frac{\sqrt{x+6}-3}{(x-3)(x+3)}=1+\frac{1}{(x+3)(\sqrt{x+6}+3)}$$

Answer 2

An elementary approach: "recognize the derivative."

You can rewrite, for $x\notin\{-3,3\}$, $$ \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} = 1+ \frac{\sqrt{x+6}-3}{x^{2}-9} = 1+ \frac{\sqrt{x+6}-3}{(x-3)(x+3)} = 1+ \frac{\sqrt{x+6}-3}{(x-3)(x+3)} \tag{1} $$ so the question boils down to computing $\lim_{x\to 3} \frac{\sqrt{x+6}-3}{x-3}$ (the rest is "under control").

Now, we can recognize a derivative here: let $f\colon[0,\infty) \to [0,\infty)$ be defined by $$ f(x) = \sqrt{x+6}, \qquad x\geq 0\,. \tag{2} $$ Note that $f$ is differentiable, with $f'(x) = \frac{1}{2\sqrt{x+6}}$. Therefore, $$ \frac{1}{6} = \frac{1}{2\sqrt{9}} = f'(3) = \lim_{x\to 3} \frac{f(x)-f(3)}{x-3} = \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x-3} \tag{3} $$ giving the desired limit.

Putting it together, from (1) and (3) we get $$ \begin{align*} \lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} &= 1+ \lim_{x\to 3} \frac{\sqrt{x+6}-3}{(x-3)(x+3)}= 1+\frac{1}{6}\lim_{x\to 3} \frac{\sqrt{x+6}-3}{x-3} \\&= 1+\frac{1}{36} = \boxed{\frac{37}{36}} \end{align*}$$

Answer 3

$$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$ $$ = \lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-9-3}{x^{2}-9}$$ $$ = \lim_{x\to 3} \frac{x^{2}-9+\sqrt{x+6}-3}{x^{2}-9}$$ $$ = \lim_{x\to 3} \Big(\frac{x^{2}-9}{x^{2}-9}+\frac{\sqrt{x+6}-3}{x^{2}-9}\Big) $$ $$ = \lim_{x\to 3} \frac{x^{2}-9}{x^{2}-9} + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$ $$ = \lim_{x\to 3} 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$ $$ = 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9}$$

After rationalizing the numerator to remove the radical: $$\lim_{x\to 3}\frac{\sqrt{x+6}-3}{x^{2}-9} \cdot \frac{\sqrt{x+6}+3}{\sqrt{x+6}+3} = \lim_{x\to 3}\frac{x+6-9}{(x^{2}-9)(\sqrt{x+6}+3)} = \lim_{x\to 3}\frac{x-3}{(x^{2}-9)(\sqrt{x+6}+3)} $$

After factoring $ x^2 - 9 $, simplifying, and direct substitution: $$ \lim_{x\to 3}\frac{x-3}{(x^{2}-9)(\sqrt{x+6}+3)} = \lim_{x\to 3}\frac{x-3}{(x-3)(x+3)(\sqrt{x+6}+3)} = \lim_{x\to 3}\frac{1}{(x+3)(\sqrt{x+6}+3)} = \frac{1}{6\cdot 6} = \frac{1}{36}$$

Thus $$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} = 1 + \lim_{x\to 3} \frac{\sqrt{x+6}-3}{x^{2}-9} = 1 + \frac{1}{36} = \frac{36}{36} + \frac{1}{36} $$ $$= \frac{37}{36}$$