3
$\begingroup$

How can I calculate limit without using L'Hopital rule?

$\displaystyle\lim_{x\to 0 }\frac{e^{\arctan(x)}-e^{\arcsin(x)}}{1-\cos^3(x)}$.

$\endgroup$
5
$\begingroup$

Let $\arctan x = t$ so that $x = \tan t$ so that $\sin t = x/\sqrt{1 + x^{2}}$ or $\arctan x = t = \arcsin (x/\sqrt{1 + x^{2}})$.

Now we have

$\displaystyle\begin{aligned}\arctan x - \arcsin x &= \arcsin (x/\sqrt{1 + x^{2}}) - \arcsin x\\ &= \arcsin\left(x\sqrt{\frac{1 - x^{2}}{1 + x^{2}}} - \frac{x}{\sqrt{1 + x^{2}}}\right)\\ &= \arcsin\left\{\frac{x}{\sqrt{1 + x^{2}}}\left(\sqrt{1 - x^{2}} - 1\right)\right\}\\ &= \arcsin\left\{\frac{-x^{3}}{\sqrt{1 + x^{2}}\left(\sqrt{1 - x^{2}} + 1\right)}\right\}\\ &= \arcsin y = f(x) \text{ (say)}\end{aligned}$

Next we have

$\displaystyle\begin{aligned}\lim_{x \to 0}\frac{e^{\arctan x} - e^{\arcsin x}}{1 - \cos^{3}x} &= \lim_{x \to 0}e^{\arcsin x}\cdot\frac{e^{f(x)} - 1}{1 - \cos^{3} x}\\ &= \lim_{x \to 0}1\cdot\frac{e^{f(x)} - 1}{f(x)}\cdot\frac{f(x)}{1 - \cos^{3}x}\\ &= \lim_{x \to 0}1\cdot 1\cdot\frac{f(x)}{(1 - \cos x)(1 + \cos x + \cos^{2}x)}\\ &= \frac{1}{3}\lim_{x \to 0}\frac{f(x)}{1 - \cos x}\\ &= \frac{1}{3}\lim_{x \to 0}\frac{\arcsin y}{y}\cdot \frac{y}{1 - \cos x}\\ &= \frac{1}{3}\lim_{x \to 0}\frac{y}{2\sin^{2}(x/2)}\\ &= \frac{1}{6}\lim_{x \to 0}\frac{y}{(x/2)^{2}}\cdot \frac{(x/2)^{2}}{\sin^{2}(x/2)}\\ &= \frac{1}{6}\lim_{x \to 0}\frac{y}{(x/2)^{2}}\cdot 1\\ &= \frac{2}{3}\lim_{x \to 0}\frac{y}{x^{2}}\\ &= \frac{2}{3}\lim_{x \to 0}\frac{-x}{\sqrt{1 + x^{2}}\left(\sqrt{1 - x^{2}} + 1\right)}\\ &= \frac{2}{3}\cdot 0 = 0\end{aligned}$

No Taylor or L'Hospital is required. We just need algebraic and trigonometric manipulation combined with the use of standard limits.

$\endgroup$
  • $\begingroup$ Could you explain me why do you think that $\lim_{x \to 0}\frac{e^{f(x)}-1}{f(x)}=1?$ Because I think we should do the following(remember that $f(x)=arctan(x)-arcsin(x)$!): $\lim_{x \to 0}\frac{e^{f(x)}-1}{f(x)}=\lim_{x \to 0}\frac{e^{arctan(x)-arcsin(x)}-1}{arctan(x)-arcsin(x)}$. And if $x \to 0$ we get $\frac{e^{0-0}-1}{0-0}=\frac{1-1}{0-0}=\frac{0}{0}$, which is indeterminate form. $\endgroup$ – k1ber Jan 13 '14 at 15:36
  • 1
    $\begingroup$ @KiberPrestupnik: please note that as $x \to 0$, The variable $z = f(x)$ also tends to zero and then we know the standard limit $$\lim_{z \to 0}\frac{e^{z} - 1}{z} = 1$$ $\endgroup$ – Paramanand Singh Jan 14 '14 at 2:06
3
$\begingroup$

Hint: remember that (from taylor series) $$e^x \sim 1 + x$$ if $x \to 0$, and $$\cos x \sim 1 - \frac{x^2}{2}$$ if $x \to 0$

Substitute into your fraction and you should get to the result easily.

The only thing that you need are the taylor expansion of arctan and arcsin

$$\arcsin x \sim x + \frac{x^3}{6}$$ $$\arctan x \sim x - \frac{x^3}{3}$$

Always with $x \to 0$

Just substitute and you're good to go

(note that this are taylor expansion limited at the third order)

(by the way, be careful with this method unless you understand what $o(x)$ means (I omitted it in my answer) and how to use it.. for example it's true that $\sin x \sim x$, but you can't write $(\sin x - x) \sim (x - x) \sim 0$, it just has no meaning. In such cases you should use the sin approximation of superior order. $\sin x \sim x - \frac{x^3}{3} \Rightarrow (\sin x - x) \sim (x - \frac{x^3}{3} - x) \sim -\frac{x^3}{3}$ )

$\endgroup$
  • $\begingroup$ arcsin x ~ x+1/6(x^3)+... $\endgroup$ – Khosrotash Jan 9 '14 at 20:34
  • $\begingroup$ you're right! Thanks :-) $\endgroup$ – Ant Jan 9 '14 at 20:35
-2
$\begingroup$

arctan x ~ x-1/3(x^3)+...
arcsin x ~ x+1/6(x^3)+...
e^x~1+x+... e^ arctan x -e^ arcsin x ~1+ x-1/3(x^3) -(1+x+1/6(x^3) )~ -3 x^3 /6
1-cos^3 x ~ (1-cos x) (1+cos x +cos^2x )~ x^2 /2 (1+cos x+cos^2x)
put them in limit
you will have lim =0

$\endgroup$
  • 2
    $\begingroup$ Please try to use LaTeX it's really not that difficult. $\endgroup$ – nbubis Jan 10 '14 at 5:14

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.