Calculate the limit without l'Hopital rule 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

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.
