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

Question

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. $0$

Answer 1

$$\begin{align*} & \lim_{x \rightarrow 0} \frac{e^x - e^{x\cos x}}{x + \sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x - 1 + 1 - e^{x\cos x}}{x + \sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x+\sin x} + \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{x+\sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot \frac{x}{x+\sin x} + \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{-x\cos x} \cdot \frac{-x\cos x}{x+\sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot \lim_{x \rightarrow 0} \frac{1}{1 + \frac{\sin x}{x}} + \lim_{x\cos x \rightarrow 0} \frac{e^{x\cos x}-1}{x\cos x} \cdot \lim_{x \rightarrow 0} -\frac{x}{x+\sin x} \cdot \cos x \\ & = 1\cdot \frac{1}{1+1} + 1 \cdot -\frac{1}{1+1} \cdot 1\\ & = \frac{1}{2} - \frac{1}{2}\\ & = 0 \\ \end{align*}$$

Answer 2

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

Answer 3

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

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

Answer 4

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

Answer 5

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 $0$.

Answer 6

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 $x\to 0$.