$\lim_{x→0}$
$\dfrac{x^2}{x\sqrt{1+x} −\ln(1+x)}= ?$
I got $-2$. Is this correct if not what is the answer so i can find out where i went wrong.
Thanks in advance
$\lim_{x→0}$
$\dfrac{x^2}{x\sqrt{1+x} −\ln(1+x)}= ?$
I got $-2$. Is this correct if not what is the answer so i can find out where i went wrong.
Thanks in advance
Using Taylor series $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+O\left(x^4\right)$$ $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right)$$ so $$x\sqrt{1+x}-\log(1+x)=x^2-\frac{11 x^3}{24}+O\left(x^4\right)$$ $$\dfrac{x^2}{x\sqrt{1+x} −\ln(1+x)}= \frac{x^2}{x^2-\frac{11 }{24}x^3+O\left(x^4\right)}=\frac{1}{1-\frac{11 }{24}x+O\left(x^2\right)}=1+\frac{11 }{24}x+O\left(x^2\right)$$ which shows the limit and also how it is approached.
To make things look nicer and easier to do, let $y = \sqrt{1+x}$, then $L = \displaystyle \lim_{y \to 1} \dfrac{(y^2-1)^2}{y(y^2-1) - 2\ln(y)}$. From this you can apply L'hospital rule much faster than the original form.