Limit of the exponential functions: $\lim_{x\to 0} \frac{e^x-e^{x \cos x}} {x +\sin x}$

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. $0$

• Are you sure it is 0? I think it is something bigger than that. Jan 12 '17 at 6:39
• I am pretty sure it is zero. I also checked on Wolfram and it gave me the same amswer. Jan 12 '17 at 6:42
• I see my problem. I had (x-sinx) in the denominator.. and the limit = 3. Jan 12 '17 at 6:44

\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*}

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

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

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 $x\to 0$.

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

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 $0$.

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

• ... 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. Jan 12 '17 at 6:27
• 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 $x \to 0$
– Fred
Jan 12 '17 at 6:38
• 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. Jan 12 '17 at 6:41
• No ! $e^x$, $\sin x$, etc... are given by power series !
– Fred
Jan 12 '17 at 6:48
• 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. Jan 12 '17 at 6:51