# Calculating limits using l'Hôpital's rule.

After a long page of solving limits using l'Hôpital's rule only those 2 left that i cant manage to solve

$$\lim\limits_{x\to0}{\sqrt {\cos x} - \sqrt{\cos x}\over \sin^2 x }$$

$$\lim\limits_{x\to\ {pi\over 2}}{\tan 3x - 3\over \tan x - 3 }$$

Thanks in advance for any help :)

i edit the second one in mistake i entered $0$ insted of $\pi\over 2$

• Since $\tan x \to 0$ as $x \to 0$ the second one can be evaluated directly using continuity. May 23, 2014 at 3:59
• Second one isn't indeterminate form! May 23, 2014 at 4:01
• For the first, note that $\sin^2 x = 1 -\cos^2 x$ and find the limit of ${ y^{1\over 2} - y^{ 1 \over 3} \over y^2 }$ as $y \to 1$ using l'Hôpital's rule. May 23, 2014 at 4:03

$$\lim\limits_{x\to0}{\sqrt {\cos x} - \sqrt{\cos x}\over \sin^2 x }=\lim\limits_{x\to0}{-\frac{1}{2}\sin x(\cos x)^{-\frac{1}{2}} + \frac{1}{3}\sin x (\cos x)^{-\frac{1}{3}}\over 2\sin x \cos x }=\lim\limits_{x\to0}{-\frac{1}{2}(\cos x)^{-\frac{1}{2}} + \frac{1}{3} (\cos x)^{-\frac{1}{3}}\over 2 \cos x }=-\frac{1}{12}$$

For the second one, you can evaluate the limit directly by substituting in $x=0$.

As $\displaystyle\sqrt{\cos x}-\sqrt{\cos x}=(\cos x)^{\frac12}-(\cos x)^{\frac13}$ and gcd$\displaystyle\left(\frac12,\frac13\right)=\frac{\text{gcd}(1,1)}{\text{lcm}(2,3)}=\frac16,$

Set $\displaystyle\sqrt{\cos x}=1+h\implies\cos x=(1+h)^6$

and $\displaystyle\sqrt{\cos x}-\sqrt{\cos x}=(1+h)^3-(1+h)^2=(1+h)^2(1+h-1)$

Finally, $\displaystyle\sin^2x=1-(1+h)^{12}=-12h+O(h^2)$

• very smart solution indeed !! May 23, 2014 at 18:37
• @ParamanandSingh, Thanks for your feedback May 24, 2014 at 12:26

@copper.hat thanks i think i got it

$$\lim\limits_{x\to0}{\sqrt {\cos x} - \sqrt{\cos x}\over \sin^2 x } = \lim\limits_{x\to0}{\sqrt {\cos x} - \sqrt{\cos x}\over 1-\cos^2 x } = \lim\limits_{y\to1}{\sqrt {y} - \sqrt{y}\over 1- y^2 }$$ $$\lim\limits_{y\to1}{\sqrt {y} - \sqrt{y}\over 1- y^2 } = \lim\limits_{y\to1}{\sqrt {y} - \sqrt{y}\over 2y(6\sqrt y) } = 0$$