2
$\begingroup$

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?

$\endgroup$
  • 7
    $\begingroup$ 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 $\endgroup$ – qbert Dec 16 '16 at 20:40
  • 2
    $\begingroup$ We could rewrite the first as $$ 1 + \lim_{x \to 0} \frac{\sin(x) - x}{x^3} $$ $\endgroup$ – Omnomnomnom Dec 16 '16 at 21:09
  • 1
    $\begingroup$ 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 $\endgroup$ – Omnomnomnom Dec 16 '16 at 21:12
  • 1
    $\begingroup$ 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
  • 4
    $\begingroup$ If you want to remove a screw, you use a screwdriver, not your finger nails. $\endgroup$ – egreg Dec 16 '16 at 21:47
3
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @VincenzoOliva Yes, I assume the limit does exist. $\endgroup$ – Olivier Oloa Dec 17 '16 at 10:26
  • 1
    $\begingroup$ Ah ok!! Thank you!! :-) $\endgroup$ – Mary Star Dec 17 '16 at 11:51
3
$\begingroup$

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 $L_2=\frac12$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

Sure: use Wolfram Alpha or Maple or ...

$\endgroup$
  • 4
    $\begingroup$ Probably should've made this a comment? $\endgroup$ – Simply Beautiful Art Dec 16 '16 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.