Calculate the limit without l'Hopital rule

Question

I have the following limit:

$$ \lim_{x\to 7}\dfrac{x^2-4x-21}{x-4-\sqrt{x+2}} $$ I could easily calculate the limit = 12 using the l'Hopital rule.

Could you please suggest any other ways to solve this limit without using the l'Hopital rule?

Thank you

Answer 1

An alternative to @MatthewDaly's comment: write $y:=\sqrt{x+2}$ so you want$$\lim_{y\to3}\frac{y^4-8y^2-9}{y^2-y-6}=\lim_{y\to3}\frac{y^3+3y^2+y+3}{y+2}=\frac{3^3+3\times 3^2+3+3}{5}=12.$$

Answer 2

$$\lim_{x→7}\frac{x^2−4x−21}{x−4−\sqrt{x+2}}\cdot\frac{x−4+\sqrt{x+2}}{x−4+\sqrt{x+2}}\\ =\lim_{x→7}\frac{(x^2−4x−21)\cdot(x−4+\sqrt{x+2})}{(x−4)^2−(x+2)}\\ =\lim_{x→7}\frac{(x^2−4x−21)\cdot(x−4+\sqrt{x+2})}{x^2-9x+14}\\ =\lim_{x→7}\frac{(x+3)\cdot(x−4+\sqrt{x+2})}{x-2}\\ =\frac{10\cdot6}{5}=12$$ since $(x-7)$ can be factored out of both of those quadratics.

Answer 3

The numerator can be factored as $(x-7)(x+3)$. Now consider $$ \lim_{x\to7}\frac{x-4-\sqrt{x+2}}{x-7}=\lim_{x\to7}\frac{x-7-(\sqrt{x+2}-3)}{x-7}= 1-\lim_{x\to7}\frac{x+2-9}{(x-7)(\sqrt{x+2}+3)}=1-\frac{1}{6}=\frac{5}{6} $$ So your limit is $$ \lim_{x\to7}\frac{x-7}{x-4-\sqrt{x+2}}(x+3)=\frac{6}{5}\cdot10=12 $$

Answer 4

Just another way to do it.

Consider $$A=\dfrac{x^2-4x-21}{x-4-\sqrt{x+2}}$$ and let $x=y+7$ to work around $y=0$. So, $$A=\frac{y (y+10)}{y+3-\sqrt{y+9}}$$ Now, using the binomial theorem or Taylor series $$\sqrt{y+9}=3+\frac{y}{6}-\frac{y^2}{216}+O\left(y^3\right)$$ making $$A=\frac{y (y+10)}{\frac{5 y}{6}+\frac{y^2}{216}+O\left(y^3\right)}=\frac{ (y+10)}{\frac{5 }{6}+\frac{y}{216}+O\left(y^2\right)}$$ Now, using long division $$A=12+\frac{17 y}{15}+O\left(y^2\right)$$ which, for sure, shows the limit but also how it is approached.