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$$L=\lim_{x\to 0}\frac{\cos(\sin x)-(1+x^2)^{\frac{-1}{2}}}{x^4}$$ To Evaluate the limit I used few terms of taylor series:

$\cos(\sin x)=1-\frac{\sin^2 x}2+\cdots$

$(1+x^2)^{\frac{-1}2}=1-\frac12x^2+\cdots$

$$L=\lim_{x\to 0}\frac{1-\frac{\sin^2 x}{2}-1+\frac{x^2}{2}}{x^4}=\lim_{x\to0}\frac{x^2-\sin^2 x}{2x^4}$$

$\sin^2 x=(x-\frac{x^3}{6}+O(x^4))^2=x^2-\frac{x^4}{3}+O(x^5)$

So we have:$$\lim_{x\to 0}\frac{x^2-x^2+\frac{x^4}{3}}{2x^4}=\frac{1}{6}$$

But the right answer is $\frac{-1}6$. Why my final answer is wrong?

Etemon
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2 Answers2

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This is a standard application of Taylor series, hence Community Wiki.

We have $\cos x=1-\frac12x^2+\frac1{24}x^4-\cdots$ and $\sin x=x-\frac16x^3+\cdots$, so $$ \begin{aligned} \cos (\sin x)&=1-\frac12\sin^2 x+\frac1{24}\sin^4x+\cdots\\ &=1-(\frac12x^2-\frac16x^4+O(x^6))+\frac1{24}(x^4+O(x^6))\\ &=1-\frac12x^2+\frac5{24}x^4+O(x^6). \end{aligned} $$ Similarly (use the binomial series) the Taylor series of $$ 1/\sqrt{1+x^2}=1-\frac12x^2+\frac9{24}x^4+O(x^6). $$ Therefore $$ \cos(\sin x)-\frac1{\sqrt{1+x^2}}=-\frac16x^4+ O(x^6). $$ From this it follows immediately that the limit is equal to $-1/6$.

Everywhere $O(x^6)$ stands for terms of degree six or higher.

The problem with the attempt in the original post is that in the first step some terms of degree four were neglected. In the end we cancel a factor of $x^4$, so this was a mistake.

Jyrki Lahtonen
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  • Thanks. I am new at using taylor series to evaluate limits. So, how many of first terms of the series should be written to get the right answer? I used very few terms and got wrong. In this case as @AdityaDwivedi pointed out we should write it until $x^4$. but how many terms needed in general? – Etemon Oct 26 '20 at 05:27
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    @soheil That depends. You see that in the end we divide by $x^4$. For that to make sense there must not be any unknown $x^4$. If, instead, we had a limit like $$\lim_{x\to0}\frac{\cos(\sin x)-1/\sqrt{1+x^2}+x^4/6}{x^6},$$ then at the end we would be cancelling a factor $x^6$. In that case it would be necessary for the series to be accurate up to terms of degree six. – Jyrki Lahtonen Oct 26 '20 at 05:31
  • Very good explanation I completely understand now. – Etemon Oct 26 '20 at 14:47
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Write $\cos(\sin x)-(1+x^2)^{1/2}$ as $\cos(\sin x)-\cos x+\cos x-(1+x^2)^{1/2}$

Then, $$\cos(\sin x)-\cos x=2\sin\dfrac{x-\sin x}2\sin\dfrac{x+\sin x}2$$

Now use L'Hospital's or Are all limits solvable without L'Hôpital Rule or Series Expansion to find $$\lim_{x\to0}\dfrac{x-\sin x}{x^3}=\dfrac16$$ so that $$\lim_{x\to0}\dfrac{\cos(\sin x)-\cos x}{x^4}=\dfrac24\lim_{x\to0}\dfrac{\sin\dfrac{x-\sin x}2}{\dfrac{x-\sin x}2}\lim_{x\to0}\dfrac{\sin\dfrac{x+\sin x}2}{\dfrac{x+\sin x}2}\lim_{x\to0}\dfrac{x-\sin x}{x^3}\lim_{x\to0}\dfrac{x+\sin x}x$$

Finally $$\cos x-(1+x^2)^{-1/2}$$

$$=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots-\left(1-\dfrac{x^2}2+x^4\cdot\dfrac{\left(-\dfrac12\right)\left(-\dfrac12-1\right)}{2!}+\cdots\right)$$

$$=x^4\cdot\left(\dfrac1{4!}-\dfrac38\right)+\text{terms containing higher exponents of }x$$

lab bhattacharjee
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