Proof by $\epsilon-\delta$ definition $\lim_{x \to 0} \frac{e^x \cos x - 1}{x} = 1$

Question

Proof by definition $\lim\limits_{x \to 0} \dfrac{e^x \cos x - 1}{x}=1$.

I could not go far because I am lacking insight of how to find $\delta>0$ in this case.

Let $\epsilon>0$.

$\displaystyle\bigg|\frac{e^x \cos x - 1}{x}-1\bigg|=\bigg|\frac{e^x\cos x-1-x}{x}\bigg|\leq\bigg|\frac{e^x-1-x}{x}\bigg|$

Any help would be appreciated.

Answer 1

$$\frac{e^x\cos(x)-1}{x} = \cos(x)\frac{e^x-1}{x}+\frac{\cos(x)-1}{x} = \cos(x)\frac{e^x-1}{x}-\frac{2\sin^2(x/2)}{x} $$ We have for some fixed $\delta$, and $0<|x|<\delta$: $$|\frac{2\sin^2(x/2)}{x}|\leq |x/2|<\delta/2 $$ For $\delta<1$ $$1+x+x^2\geq e^x\geq 1+x \implies |\frac{e^x-1}{x}-1|\leq|x|<\delta$$ $$ $$

$$|\frac{e^x\cos(x)-1}{x}-1|\leq |1-\cos(x)|+ |\frac{e^x-1}{x}-1|+|\frac{2\sin^2(x)}{x}|< \delta^2+2\delta $$ $$ \varepsilon = \delta^2+2\delta \implies \delta = \sqrt{\varepsilon+1}-1 $$ We went backwards, but it's the final result that matters.

For any $\varepsilon>0$, if we take $\delta = \min{\{\sqrt{\varepsilon+1}-1, 1\}}>0$, then$$0<|x|<\delta \implies |\frac{e^x\cos(x)-1}{x}-1|< \varepsilon $$ From the definition of a limit, $\lim_{x\to 0} \frac{e^x\cos(x)-1}{x} = 1$

Answer 2

You have an indetermined expression "0/0" involving smooth numerator and denominator. So you can perform Taylor expansions of both, and first order suffices. This way $\exp(x) \simeq 1+x$, $\cos x \simeq 1$, and the numerator $\simeq x$. The limit follows.

Answer 3

By definition of the derivative one has

$$\lim_{ x \to 0 } \frac{ e^x \cos(x) - 1 }{x}$$

$$= \lim_{ x \to 0 } \frac{ e^x \cos(x) - e^0 cos(0) }{x - 0}$$

$$= \frac{d}{dx} \Big|_{x=0} ( e^x \cos(x) )$$

$$= e^x ( \cos(x) - \sin(x) ) \Big|_{x=0}$$

$$= 1$$

And we did not use L'Hospital.

Answer 4

Notice that $$0\leq e^x-1-x \leq x^2,~~~|x|\leq 1.$$ and $$0\leq \sin^2 x \leq x^2,~~~x \in \mathbb{R}.$$

Let's say that $|x|\leq 1.$ Hence,

\begin{align*}\left|\frac{e^x\cos x-1}{x}-1\right|&=\left|\frac{e^x\left(1-2\sin^2\dfrac{x}{2}\right)-1-x}{x}\right|\\&\leq\left|\frac{e^x-1-x}{x}\right|+\left|\frac{e^x\cdot2\sin^2\dfrac{x}{2}}{x}\right|\\&\leq\frac{x^2}{|x|}+\frac{e^x\cdot 2\cdot(\dfrac{x}{2})^2}{|x|}\\&=\left(1+\frac{e}{2}\right)|x|.\end{align*}

Now, let $$\left(1+\frac{e}{2}\right)|x|<\varepsilon.$$ We obtain $$|x|<\dfrac{\varepsilon}{1+\dfrac{e}{2}}.$$

Thus, we take $$\delta=\min\left(1,\dfrac{\varepsilon}{1+\dfrac{e}{2}}\right),$$then when $|x|<\delta$, we have $$\left|\frac{e^x\cos x-1}{x}-1\right|<\varepsilon.$$