# Find a limit of a function W/OUT l'Hopital's rule.

I've got an expression: $\lim_{x\to 0}$ $\frac {log(6-\frac 5{cosx})}{\sin^2 x}$

The question is: how to find limit without l'Hopital's rule?

Hint:

$$\dfrac{\ln\left(6-\dfrac5{\cos x}\right)}{\sin^2x}=\dfrac{\ln(6\cos x-5)}{\sin^2x}+\dfrac{\ln(1-\sin^2x)}{-2\sin^2x}$$

Now the second limit can be managed by $\lim_{h\to}\dfrac{\ln(1+h)}h=1$

For the first limit $6\cos x-5=6\left(1-2\sin^2\dfrac x2\right)-5=1-12\sin^2\dfrac x2$

and $\sin^2x=4\sin^2\dfrac x2\cos^2\dfrac x2$

Another way using Taylor series $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^5\right)$$ $$\frac 1 {\cos(x)}=1+\frac{x^2}{2}+\frac{5 x^4}{24}+O\left(x^5\right)$$ $$6-\frac 5 {\cos(x)}=1-\frac{5 x^2}{2}-\frac{25 x^4}{24}+O\left(x^5\right)$$ $$\log\left(6-\frac 5 {\cos(x)} \right)=-\frac{5 x^2}{2}-\frac{25 x^4}{6}+O\left(x^5\right)$$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\sin^2(x)=x^2-\frac{x^4}{3}+O\left(x^6\right)$$ $$\frac{\log\left(6-\frac 5 {\cos(x)} \right) } {\sin^2(x) }=\frac{-\frac{5 x^2}{2}-\frac{25 x^4}{6}+O\left(x^5\right) }{x^2-\frac{x^4}{3}+O\left(x^6\right) }=-\frac{5}{2}-5 x^2+O\left(x^3\right)$$ which shows the limit and also how it is approached.

• Haven't heard of it before because I'm only first grade and aout professor is gonna tell us about this method in a few weeks but also thank you! – Эмиль Бураиндов Dec 7 '17 at 7:18