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I want to find the limit of this function by simply using algebraic manipulation. Though I have computed the limit through L' Hospital's method but still I want to compute the limit purely by function's manipulation to yield a form where limit can be applied $$\lim_{x\to 0} \frac{e^x-e^{x \cos x}} {x +\sin x} $$

Till now we have been taught basic limits such as $\lim_{x\to 0} \frac{e^x-1}{x}=1$ and that's why I have been trying to bring such form in this expression.

P.S. I got the current answer by L'Hospital's rule i.e. $0$

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  • $\begingroup$ Are you sure it is 0? I think it is something bigger than that. $\endgroup$
    – Doug M
    Jan 12 '17 at 6:39
  • $\begingroup$ I am pretty sure it is zero. I also checked on Wolfram and it gave me the same amswer. $\endgroup$ Jan 12 '17 at 6:42
  • $\begingroup$ I see my problem. I had (x-sinx) in the denominator.. and the limit = 3. $\endgroup$
    – Doug M
    Jan 12 '17 at 6:44
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$$\begin{align*} & \lim_{x \rightarrow 0} \frac{e^x - e^{x\cos x}}{x + \sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x - 1 + 1 - e^{x\cos x}}{x + \sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x+\sin x} + \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{x+\sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot \frac{x}{x+\sin x} + \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{-x\cos x} \cdot \frac{-x\cos x}{x+\sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot \lim_{x \rightarrow 0} \frac{1}{1 + \frac{\sin x}{x}} + \lim_{x\cos x \rightarrow 0} \frac{e^{x\cos x}-1}{x\cos x} \cdot \lim_{x \rightarrow 0} -\frac{x}{x+\sin x} \cdot \cos x \\ & = 1\cdot \frac{1}{1+1} + 1 \cdot -\frac{1}{1+1} \cdot 1\\ & = \frac{1}{2} - \frac{1}{2}\\ & = 0 \\ \end{align*}$$

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$$\frac{e^x-e^{x \cos x}} {x +\sin x}=\frac{\frac{e^x}{x}-\frac{e^{x \cos x}}{x}}{1+\frac{\sin x}{x}}=\frac{\frac{e^x-1}{x}-\frac{e^{x \cos x}-1}{x}}{1+\frac{\sin x}{x}}=\frac{\frac{e^x-1}{x}-\cos x\frac{e^{x \cos x}-1}{x \cos x}}{1+\frac{\sin x}{x}}$$

Hence the limit is $\frac{1-1(1)}{1+1}=0$.

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Note that $$\frac{e^{x} - e^{x\cos x}} {x+\sin x} = e^{x\cos x} \cdot\frac{e^{x(1-\cos x)} - 1}{x(1-\cos x)} \cdot\dfrac{1-\cos x} {1 + \dfrac{\sin x} {x}} $$ which tends to $$e^{0\cdot 1}\cdot 1\cdot\frac{1-1}{1+1}=0$$ as $x\to 0$.

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$$\lim _{x\to 0}\left(\frac{e^x-e^{x\:\cos(x)}}{\:x+sin(x)}\:\right) \approx \lim _{x\to 0}\left(\frac{1+x+o\left(x^2\right)-\left(1+x+o\left(x^2\right)\right)}{\:x+x+o\left(x^3\right)}\:\right) = \color{red}{0}$$

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Consider $$ \frac{e^x-e^{x\cos x}}{x+\sin x}=\frac{e^x(1-e^{x(\cos x-1)})}{x+\sin x}= \frac{-xe^x}{x+\sin x}\frac{e^{x(\cos x-1)}-1}{x} $$ The first factor is easily seen to have limit $-1/2$. The limit of the second factor is $$ \lim_{x\to0}\frac{e^{x(\cos x-1)}-1}{x} $$ which is the derivative at $0$ of $f(x)=e^{x(\cos x-1)}$. Any manipulation you do will just be adapting the proof of the chain rule to this particular case, so why not using it directly? Since $$ f'(x)=e^{x(\cos x-1)}(\cos x-1-x\sin x) $$ we have $f'(0)=0$.

Without the derivative, $$ \lim_{x\to0}\frac{e^{x(\cos x-1)}-1}{x}= \lim_{x\to0}\frac{e^{x(\cos x-1)}-1}{x(\cos x-1)}(\cos x-1) $$ and the limit of the fraction is $1$ because it's essentially the same as $$ \lim_{t\to0}\frac{e^t-1}{t} $$ because $\lim_{x\to0} x(\cos x-1)=0$ and the function $g(x)=x(\cos x-1)$ is invertible in a neighborhood of $0$.

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Hint: series expansion of $e^x$, $e^{x \cos x}$ and $\sin x$.

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    $\begingroup$ ... That's L'Hopital in a nutshell. Or L'Hopital is this in a nutshell. Or something. Either way, differentiating the numerator and denominator until you get non-zero terms is what the OP wants to avoid if I read him correctly. $\endgroup$
    – Arthur
    Jan 12 '17 at 6:27
  • $\begingroup$ Thats not true !! The method of computing limits with power series only needs the continuity of power series on the open intervall of convegence. No derivatives , etc .. are needed. Example: for $x \ne 0$: $\frac{e^x-1}{x}=1+\frac{x}{2!}+\frac{x^2}{3!}+... \to 1$ for $x \to 0$ $\endgroup$
    – Fred
    Jan 12 '17 at 6:38
  • $\begingroup$ You use differentiation to find the coefficients of the power series of the numerator and denominator in the first place, before you carry out the division. $\endgroup$
    – Arthur
    Jan 12 '17 at 6:41
  • $\begingroup$ No ! $e^x$, $ \sin x$, etc... are given by power series ! $\endgroup$
    – Fred
    Jan 12 '17 at 6:48
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    $\begingroup$ Not always. The are plenty of other ways to define them both. Unless you know how OP has defined them, you can't assume it's through power series. $\endgroup$
    – Arthur
    Jan 12 '17 at 6:51

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