Vtyjkyuk

Proof by definition $limlimits_x to 0 dfrace^x cos x - 1x=1$.

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

Let $epsilon>0$.

$displaystylebigg|frace^x cos x - 1x-1bigg|=bigg|frace^xcos x-1-xxbigg|leqbigg|frace^x-1-xxbigg|$

Any help would be appreciated.

$$frace^xcos(x)-1x = cos(x)frace^x-1x+fraccos(x)-1x = cos(x)frace^x-1x-frac2sin^2(x/2)x $$
We have for some fixed $delta$, and $0<|x|<delta$:
$$|frac2sin^2(x/2)x|leq |x/2|<delta/2 $$
For $delta<1$
$$1+x+x^2geq e^xgeq 1+x implies |frace^x-1x-1|leq|x|<delta$$
$$ $$

$$|frace^xcos(x)-1x-1|leq |1-cos(x)|+ |frace^x-1x-1|+|frac2sin^2(x)x|< delta^2+2delta $$
$$ varepsilon = delta^2+2delta implies delta = sqrtvarepsilon+1-1 $$
We went backwards, but it's the final result that matters.

For any $varepsilon>0$, if we take $delta = minsqrtvarepsilon+1-1, 1>0$, then$$0<|x|<delta implies |frace^xcos(x)-1x-1|< varepsilon $$
From the definition of a limit, $lim_xto 0 frace^xcos(x)-1x = 1$

By definition of the derivative one has

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

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

$$= fracddx Big|_x=0 ( e^x cos(x) )$$

$$= e^x ( cos(x) - sin(x) ) Big|_x=0$$

$$= 1$$

And we did not use L'Hospital.

Notice that $$0leq e^x-1-x leq x^2,~~~|x|leq 1.$$ and $$0leq sin^2 x leq x^2,~~~x in mathbbR.$$

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

beginalign*left|frace^xcos x-1x-1right|&=left|frace^xleft(1-2sin^2dfracx2right)-1-xxright|\&leqleft|frace^x-1-xxright|+left|frace^xcdot2sin^2dfracx2xright|\&leqfracx^2+frace^xcdot 2cdot(dfracx2)^2\&=left(1+frace2right)|x|.endalign*

Now, let $$left(1+frace2right)|x|<varepsilon.$$ We obtain $$|x|<dfracvarepsilon1+dfrace2.$$

Thus, we take $$delta=minleft(1,dfracvarepsilon1+dfrace2right),$$then when $|x|<delta$, we have $$left|frace^xcos x-1x-1right|<varepsilon.$$

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.