# How to calculate limit without L'Hopital

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

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

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.

• 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. Commented Jan 13, 2014 at 15:36
• @KiberPrestupnik: please note that as $x \to 0$, The variable $z = f(x)$ also tends to zero and then we know the standard limit $$\lim_{z \to 0}\frac{e^{z} - 1}{z} = 1$$ Commented Jan 14, 2014 at 2:06

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. $\sin x \sim x - \frac{x^3}{3} \Rightarrow (\sin x - x) \sim (x - \frac{x^3}{3} - x) \sim -\frac{x^3}{3}$ )

• arcsin x ~ x+1/6(x^3)+... Commented Jan 9, 2014 at 20:34
• you're right! Thanks :-)
– Ant
Commented Jan 9, 2014 at 20:35

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

• Please try to use LaTeX it's really not that difficult. Commented Jan 10, 2014 at 5:14