$$ \lim_{x\to 0}\frac{(1+\sin x)^{\operatorname{cosec}x} - e + \left(\dfrac{\sin x}{2}\right)e}{\sin^2x} $$
I am stucked here , please tell me how to proceed further and Is there any way to solve this problem
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HINT:
Recall that $e^x=1+x+\frac12 x^2 +\cdots$
SPOILER ALERT: Scroll over the highlighted area to reveal the solution
Starting with the limit $L$ as given by $$\begin{align}L&=e\lim_{x\to 0}\frac{e^{-\frac12 x+\frac13x^3-\cdots}\left(1-\frac12 x\right)}{x^2}\\\\&=e\lim_{x\to 0} \frac{1-\frac12 x+\frac13 x^2 +\frac12\left(-\frac12 x\right)^2+O\left(x^3\right)-\left(1-\frac12 x\right)}{x^2}\\\\&=\frac{11}{24}e\end{align}$$
NOTE:
Another approach is to use L'Hospital's Rule to write
$$\lim_{x\to 0}\frac{(1+x)^{1/x}-e+e x/2}{x^2}=\frac12\lim_{x\to 0}\left(\frac{d^2\,\left(1+x\right)^{1/x}}{dx^2}\right)$$
and carry out the differentiation and the limit. However, this approach does not increase efficiency over the method in the OP.
Mark Viola
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Is there an easier method – Aakash Kumar Jun 12 '16 at 13:53
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@Dr.MV I wonder how is second step correct i.e by taking limit of exponent and what is inside the bracket. How is this legit? thanks – Gathdi Jun 12 '16 at 13:56
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You're welcome. My pleasure. – Mark Viola Jun 12 '16 at 14:01
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@Gathdi There were several issues with the development that are not quite rigorous, but do provide the correct result. If one lets $y=\sin(x)$, then the steps in question are rigorous taking the limit as $y\to 0$. – Mark Viola Jun 12 '16 at 14:03
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1@user346279 I don't know if it is markedly easier, but if one begins by substituting $y=\sin(x)$, and take the limit as $y\to 0$, we arrive immediately at the third line in the picture embedded in the OP. – Mark Viola Jun 12 '16 at 14:10
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@Dr.MV hmm thanks – Gathdi Jun 12 '16 at 14:39
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@Gathdi You're welcome. My pleasure. -Mark – Mark Viola Jun 12 '16 at 14:51
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Your working has a flaw which BTW does not lead to an incorrect answer and hence it is difficult to understand that it is a flaw. While going from your second step to third step you have replaced the expression $(\sin x)/x$ with $1$. This is valid (and common practice) for the replacement done in denominator, but all the replacements (there are three) in the numerator are invalid. There is no specific theorem which guarantees the validity of such replacements in general. The result is correct however as can be seen below.
We can proceed in the following manner \begin{align} L &= \lim_{x \to 0}\frac{(1 + \sin x)^{\operatorname{cosec}x} - e + \left(\dfrac{\sin x}{2}\right)e}{\sin^{2}x}\notag\\ &= \lim_{x \to 0}\dfrac{(1 + \sin x)^{1/\sin x} - e + (e\sin x)/2}{\sin^{2}x}\notag\\ &= \lim_{t \to 0}\dfrac{(1 + t)^{1/t} - e + te/2}{t^{2}}\text{ (putting }t = \sin x)\notag\\ \end{align} You can then proceed easily by applying taylor series for $(1 + t)^{1/t}$ (like in the answer by Dr. MV).
Paramanand Singh
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It is mandatory that it must be in denominator for Sinx/x to be 1 ? Please tell me where I should use this limit . – Aakash Kumar Jun 13 '16 at 16:50
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@user346279: No. Please see this answer (http://math.stackexchange.com/a/1783818/72031) for more details on when we can do such replacements. – Paramanand Singh Jun 13 '16 at 18:25