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Could we compute the limits $$\lim_{x\rightarrow 0}\frac{\sin (x)-x+x^3}{x^3} \\ \lim_{x\rightarrow 0}\frac{e^x-\sin (x)-1}{x^2}$$ without using the l'Hospital rule and the Taylor expansion?

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    $\begingroup$ just out of curiosity, why do you want to do this without the two tools that would make this a very easy exercise? The first one is begging you to taylor expand $\endgroup$ – operatorerror Dec 16 '16 at 20:40
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    $\begingroup$ We could rewrite the first as $$ 1 + \lim_{x \to 0} \frac{\sin(x) - x}{x^3} $$ $\endgroup$ – Ben Grossmann Dec 16 '16 at 21:09
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    $\begingroup$ If we could deduce that $$ \lim_{x\to 0} \frac{\frac{\sin(x)}{x} - 1}{\cos(x) - 1} = 1 $$ that would get us most of the way there for the first problem $\endgroup$ – Ben Grossmann Dec 16 '16 at 21:12
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    $\begingroup$ According to my dictionary, compute means I am to calculate either through mathematical means or via a computer/calculator. Taking the second option, these limits aren't very hard to calculate. $\endgroup$ – Simply Beautiful Art Dec 16 '16 at 21:46
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    $\begingroup$ If you want to remove a screw, you use a screwdriver, not your finger nails. $\endgroup$ – egreg Dec 16 '16 at 21:47
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Hint. If one recalls that, for any function $f$ differentiable near $0$, $$ \lim_{x \to 0}\frac{f(x)-f(0)}{x-0}=f'(0) \tag1 $$ then one may write $$ \begin{align} l:=\lim_{x \to 0}\frac{e^x-\sin (x)-1}{x^2}=\lim_{x \to 0}\:\frac1x\left(\frac{e^x-1}{x} -\frac{\sin x}x\right) \end{align} $$ then one may apply $(1)$ to $$ f(x)=\frac{e^x-1}{x} -\frac{\sin x}x,\quad f(0)=0, $$$$ f'(x)=-\frac{e^x-\sin (x)-1}{x^2}+\frac{e^x}{x} -\frac{\cos x}x, $$ getting $$ l=-l+\lim_{x \to 0}\frac{e^x-1}{x}-\lim_{x \to 0}\frac{\cos x-1}{x} $$ and $$ l=-l+1-0 $$ that is

$$ l=\lim_{x \to 0}\frac{e^x-\sin (x)-1}{x^2}=\frac12. $$

Can you take it from here applying it to the first limit?

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    $\begingroup$ You're assuming $f$ is differentiable at $0$, i.e. that $f'(0)=l$ exists, right? $\endgroup$ – Vincenzo Oliva Dec 17 '16 at 9:46
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    $\begingroup$ @VincenzoOliva Yes, I assume the limit does exist. $\endgroup$ – Olivier Oloa Dec 17 '16 at 10:26
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    $\begingroup$ Ah ok!! Thank you!! :-) $\endgroup$ – Mary Star Dec 17 '16 at 11:51
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The first one is equal to $1+\lim\limits_{x\to0}\dfrac{\sin x - x}{x^3}.$ We can apply to this limit the technique in this answer of mine.

Let $L=\lim\limits_{x\to0}\dfrac{\sin x - x}{x^3}$. This is not infinite if it exists because near $0$ one has $$-\frac12\leftarrow\frac{\cos x-1}{x^2}<\frac{\frac{\sin x}{x}-1}{x^2}<\frac{1-1}{x^2}=0.$$ Then, considering a similar limit, $$\lim_{x\to0}\frac{\sin x\cos x-x}{x^3}=\lim_{x\to0}\frac{\sin(2x)-2x}{2x^3}=4L$$ we can deduce$$\lim_{x\to0}\frac{\sin x\cos x-\sin x}{x^3}=\lim_{x\to0}\frac{\sin x}{x}\cdot\lim_{x\to0}\frac{\cos x-1}{x^2}=-\frac12=3L,$$that is $L=-\frac16.$ So your limit equals $\frac56$.

Getting back to this; due to the finiteness of the first limit, the second one is the same as $$\lim_{x\to0}\frac{e^x-1-x}{x^2}+\lim_{x\to0}\frac{x-\sin x}{x^2}=\lim_{x\to0}\frac{\frac{e^x-1}{x}-1}{x},$$i.e. $f'(0)$ where $f(x)=\frac{e^x-1}{x}, f(0)=1.$ One has $$f'(0)=\lim_{x\to0}\frac{e^x(x-1)+1}{x^2}=\lim_{x\to0}\frac{e^x-1}{x}-\lim_{x\to0}\frac{e^x-1-x}{x^2}=1-f'(0),$$ so it equals $\frac12$.

Alternatively, let $L_2$ be this limit. It is largely known that $e^x>1+x$; this also means $e^{-x}>1-x$, or equivalently for $x<1$, $e^x<\frac1{1-x},$ and it's easy to prove $\frac1{1-x}<1+x+2x^2 $ for $0\ne x<\frac12.$ Thus, $0\le L_2\le2.$ Assuming it exists, we have $$L_2=\lim_{x\to0}\frac{e^x-1-x}{x^2}=\lim_{x\to0}\frac{e^{2x}-(1+x)^2}{x^2(e^x+1+x)}=\frac12\lim_{x\to0} \frac{e^{2x}-1-2x-x^2}{x^2}=2L_2-\frac12,$$whence $L_2=\frac12$.

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  • $\begingroup$ Why does it hold that $-\frac12\leftarrow\frac{\cos x-1}{x^2}$ ? $\endgroup$ – Mary Star Dec 17 '16 at 11:52
  • $\begingroup$ @Mary: Multiply top and bottom by $\cos x +1$ to get $$\frac{\cos^2x-1}{x^2(\cos x+1)}=-\frac{\sin^2x}{x^2}\cdot\frac{1}{\cos x+1}.$$ $\endgroup$ – Vincenzo Oliva Dec 17 '16 at 12:41
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Sure: use Wolfram Alpha or Maple or ...

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    $\begingroup$ Probably should've made this a comment? $\endgroup$ – Simply Beautiful Art Dec 16 '16 at 21:40

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