Find the limit as $x$ approaches $9$

Question

$f(x) = \dfrac{\sqrt{1+\sqrt{x}} - \sqrt{2x - 9} + 1}{x^2 - 81}$

Find the limit as x approaches 9. I've been attempting to find an anti-derivative, $g(x)$ such that the limit as x approaches 9 of $f(x)$ is equal to $g'(x)$.

Answer 1

Hint:   write it as $\,\dfrac{1}{x+9}\left(\dfrac{\sqrt{1+\sqrt{x}} - 2}{x - 9} - \dfrac{\sqrt{2x - 9} -3}{x - 9}\right)\,$ and note that each term has a limit, which doesn't even require l'Hopital to determine.

Answer 2

You're after the limit$$\lim_{x\to9}\frac{\sqrt{1+\sqrt x}-\sqrt{2x-9}+1}{(x-9)(x+9)}.$$So if $g(x)=\sqrt{1+\sqrt x}-\sqrt{2x-9}$, the number that you're after is $\frac{g'(9)}{18}$.

Answer 3

To make life simpler, start using $x=y+9$ to make $$ \dfrac{\sqrt{1+\sqrt{x}} - \sqrt{2x - 9} + 1}{x^2 - 81}=\frac{\sqrt{1+\sqrt{y+9}}- \sqrt{2 y+9}+1}{y(18+y)}$$ Now, using Taylor or the binomial theorem

$$\sqrt{y+9}=3+\frac{y}{6}-\frac{y^2}{216}+O\left(y^3\right)$$ $$\sqrt{1+\sqrt{y+9}}=\sqrt{4+\frac{y}{6}-\frac{y^2}{216}+O\left(y^3\right)}=2+\frac{y}{24}-\frac{11 y^2}{6912}+O\left(y^3\right)$$ $$\sqrt{2 y+9}=3+\frac{y}{3}-\frac{y^2}{54}+O\left(y^3\right)$$ making $$\frac{\sqrt{1+\sqrt{y+9}}- \sqrt{2 y+9}+1}{y(18+y)}=\frac{-\frac{7 y}{24}+\frac{13 y^2}{768}+O\left(y^3\right) }{y(18+y) }=\frac{-\frac{7 }{24}+\frac{13 y}{768}+O\left(y^2\right) }{(18+y) }$$ which shows the limit when $y\to 0$.

If you perform the long division, you should get $$\frac{\sqrt{1+\sqrt{y+9}}- \sqrt{2 y+9}+1}{y(18+y)}=-\frac{7}{432}+\frac{229 y}{124416}+O\left(y^2\right)$$ which also shows how the limit is approached.