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The problem is $$\lim_{x\to 0} \frac{(x \sqrt{1 + \sin x} - \ln{\sqrt{(1 + x^2)}-x)}}{\tan^3{x}} $$

I know that $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}... $$ got this $$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15}...$$

and suppose to use taylor expansion for $\ln {(1 + x)}$

Tried to expand $\sqrt{1 + \sin x}$ and $\ln{\sqrt{(1 + x^2)}}$ but it seems to be messy. Am I on the right direction?

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    Do you know asymptotic method? namely the Big O method? that method is suitable for solving your question. – 2017-02-05
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    @DeepSea Don't know that method. Can you show it? – 2017-02-05
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    @DeepSea Sorry, the initial statement was wrong. Just edited it – 2017-02-05

2 Answers 2

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$$\ln\sqrt{1+x^2}=\dfrac{\ln(1+x^2)}2=\dfrac12\left(x^2-\dfrac{x^4}2+\dfrac{x^6}3-\cdots\right)$$

Now $1+\sin x=\left(\cos\dfrac x2+\sin\dfrac x2\right)^2$

Now $\cos\dfrac x2+\sin\dfrac x2=\sqrt2\sin\left(\dfrac x2+\dfrac \pi4\right)>0$ for $x\to0$

$\implies\sqrt{1+\sin x}=\cos\dfrac x2+\sin\dfrac x2$

and $\cos\dfrac x2=1-\dfrac{(x/2)^2}{2!}+\dfrac{(x/2)^2}{4!}-\cdots$

$\sin\dfrac x2=x/2-\dfrac{(x/2)^3}{3!}+\dfrac{(x/2)^5}{5!}-\cdots$

Dord it help?

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    Thanks a lot. Understood completely. – 2017-02-05
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    @Sergei, to show appreciation, you should upvote answers that you like, and accept the one you find most satisfying. It's all explained at the [Tour](http://math.stackexchange.com/tour) of the site, really. – 2017-02-05
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Note that $$x\sqrt{1+\sin x} = x\sqrt{1+x+O(x^3)} = x(1+\frac x2-\frac{x^2}8+O(x^3)) = x+\frac{x^2}{2}-\frac{x^3}8+O(x^4)$$

and

$$\ln\sqrt{1+x^2} = \ln(1+\frac{x^2}2+O(x^4)) = \frac{x^2}{2}+O(x^4)$$

so we have

$$\lim_{x\to 0}\frac{x\sqrt{1+\sin x}-\ln\sqrt{1+x^2}-x}{\tan^3x} = \lim_{x\to 0}\frac{-\frac{x^3}8+O(x^4)}{x^3+O(x^4)} = -\frac 18$$