Question

\[\Huge{\color{red}{\lim_{x \rightarrow 0}\frac{\cos(\sin x)-cosx}{x^4}}}\]

Answer 1

\[\Huge{\color{blue}{a).1}}\]\[\Huge{\color{blue}{b).6}}\]\[\Huge {\color{blue}{c).\frac{-1}{6}}}\]\[\Huge{\color{blue}{d).\frac{1}{6}}}\]

Answer 2

this is an indeterminate 0/0 form. Have you tried using L'Hopital's Rule?

Answer 3

It is a bit bulky so I used wolfram alpha to evaluate the limit

https://www.wolframalpha.com/input/?i=limit(x-%3E0)+(cos(sinx)-cosx)%2Fx%5E4

You can try substituting in small values of x close to 0 on either side and seeing how it converges.

Answer 4

without using L Hospital rule, can u solve it ?

Answer 5

@zepdrix

Answer 6

My approach without using L'Hospital rule\[\Large \lim_{x \rightarrow 0}\frac{\cos(sinx)-cosx}{x^4}\]\[\Large{\lim_{x \rightarrow 0}\frac{-2\sin \left( \frac{sinx+x}{2} \right)\sin\left( \frac{sinx-x}{2} \right)}{x^4}}\]

Answer 7

Via series:\[\begin{align*}
\cos x&=\sum_{n\ge0}\frac{(-1)^n x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\\[1ex]
\sin x&=\sum_{n\ge0}\frac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\\[1ex]
\cos(\sin x)&=\sum_{n\ge0}\frac{(-1)^n(\sin x)^{2n+1}}{(2n+1)!}\\[1ex]
&=1-\frac{\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^2}{2!}\\[1ex]
&\quad\quad+\frac{\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^4}{4!}\\[1ex]
&\quad\quad-\frac{\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^6}{6!}+\cdots\\[1ex]
&=1-\left(\frac{x^2}{2}-\frac{x^4}{6}+\frac{x^6}{45}-\cdots\right)+\left(\frac{x^4}{24}-\frac{x^6}{36}+\cdots\right)-\left(\frac{x^6}{720}-\cdots\right)\\[1ex]
&=1-\frac{x^2}{2}+\frac{5x^4}{24}-\frac{37x^6}{720}+\cdots
\end{align*}\]Then the limit is
\[\lim_{x\to0}\frac{\left(1-\frac{x^2}{2}+\frac{5x^4}{24}-\frac{37x^6}{720}+\cdots\right)-\left(1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\cdots\right)}{x^4}\\[1ex]
=\lim_{x\to0}\frac{\frac{x^4}{6}-\frac{13x^6}{360}+\cdots}{x^4}=\lim_{x\to0}\left(\frac{1}{6}-\frac{13x^2}{360}+\cdots\right)=\frac{1}{6}\]

Answer 8

thanx a lot