How many times must I apply L’Hopital?

Question

I have this limit:

$$\lim _{x\to 0}\left(\frac{e^{x^2}+2\cos \left(x\right)-3}{x\sin \left(x^3\right)}\right)=\left(\frac 00\right)=\lim _{x\to 0}\frac{\frac{d}{dx}\left(e^{x^2}+2\cos \left(x\right)-3\right)}{\frac{d}{dx}\left(x\sin \left(x^3\right)\right)}$$ $$\lim_{x\to0}\frac{2e^{x^2}x-2\sin \left(x\right)}{\frac{d}{dx}\left(x\right)\sin \left(x^3\right)+\frac{d}{dx}\left(\sin \left(x^3\right)\right)x}=\lim_{x\to0}\frac{2e^{x^2}x-2\sin \left(x\right)}{\sin \left(x^3\right)+3x^3\cos \left(x^3\right)+\sin \left(x^3\right)}$$ but yet we have $(0/0)$. If I apply L’Hopital again, I obtain $$=\lim_{x\to0}\frac{2\left(2e^{x^2}x^2+e^{x^2}\right)-2\cos \left(x\right)}{15x^2\cos \left(x^3\right)-9x^5\sin \left(x^3\right)}$$ again giving $(0/0)$. But if I apply L’Hopital a thousand times I'll go on tilt. What is the best solution in these cases? With the main limits or applying upper bonds?

Answer 1

I'll keep this at high school level, without Taylor expansion that's quite an advanced topic. So just l'Hôpital and standard limits.

Suppose you have to compute a limit of the form $$ \lim_{x\to c}\frac{f(x)}{g(x)}\tag{*} $$ and that you know that, for some function $h$, you also have $$ \lim_{x\to c}\frac{h(x)}{g(x)}=1 $$ then you can rewrite (*) as $$ \lim_{x\to c}\frac{f(x)}{h(x)}\frac{h(x)}{g(x)} $$ As a consequence of the theorems on limits, if $$ \lim_{x\to c}\frac{f(x)}{h(x)}=l $$ then also the limit (*) will be $l$ as well.

In your case you can note that $$ 1=\lim_{x\to0}\frac{x^3}{\sin(x^3)}=\lim_{x\to0}\frac{x^4}{x\sin(x^3)} $$ and so you can reduce to computing $$ \lim_{x\to0}\frac{e^{x^2}+2\cos x-3}{x^4} $$ that's much less demanding in terms of l'Hôpital. Let's apply it to get $$ \lim_{x\to0}\frac{2xe^{x^2}-2\sin x}{4x^3}=\lim_{x\to0}\frac{xe^{x^2}-\sin x}{2x^3}\tag{**} $$ Now, what's the limit I saw some days ago with the sine and $x^3$? Yes, that one! $$ \lim_{x\to0}\frac{x-\sin x}{x^3}=\frac{1}{6} $$ OK, let's subtract and add $x$ in the numerator of (**): $$ \lim_{x\to0}\frac{xe^{x^2}-x+x-\sin x}{2x^3}= \lim_{x\to0}\frac{1}{2}\left(\frac{e^{x^2}-1}{x^2}+\frac{x-\sin x}{x^3}\right) =\frac{1}{2}\left(1+\frac{1}{6}\right)=\frac{7}{12} $$

Answer 2

You have one too many copies of the $\sin x^3$ in your denominator after the first derivative, so let's take a step back. If you're prepared to use Taylor series,$$\begin{align}e^{x^2}+2\cos x-3&=1+x^2+\frac12x^4+2-x^2+\frac{1}{12}x^4-3+o(x^4)\\&\sim\frac{7}{12}x^4,\\x\sin x^3&\sim x^4,\end{align}$$so the limit is $\frac{7}{12}$.

Answer 3

Let's first attack the numerator alone, repetitively differentiating until we no longer get zero. Let $N = \mathrm{e}^{x^2} + 2 \cos x - 3$. \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} N &= 2 x \mathrm{e}^{x^2} - 2 \sin x \xrightarrow{x \rightarrow 0} 0 \text{,} \\ \frac{\mathrm{d}^2}{\mathrm{d}x^2} N &= (4 x^2 +2)\mathrm{e}^{x^2} - 2 \cos x \xrightarrow{x \rightarrow 0} 0 \text{,} \\ \frac{\mathrm{d}^3}{\mathrm{d}x^3} N &= (8 x^3 +12 x)\mathrm{e}^{x^2} + 2 \sin x \xrightarrow{x \rightarrow 0} 0 \text{,} \\ \frac{\mathrm{d}^4}{\mathrm{d}x^4} N &= (16 x^4 +48x^2+12)\mathrm{e}^{x^2} + 2 \cos x \xrightarrow{x \rightarrow 0} 14 \text{.} \end{align*} Now let $D = x \sin x^3$. If any of its first three derivatives are nonzero, our limit is zero. Otherwise, the fourth derivative will resolve the value of the limit. \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} D &= 3 x^3 \cos x^3 + \sin x^3 \xrightarrow{x \rightarrow 0} 0 \text{,} \\ \frac{\mathrm{d}^2}{\mathrm{d}x^2} D &= 12 x^2 \cos x^3 - 9x^5 \sin x^3 \xrightarrow{x \rightarrow 0} 0 \text{,} \\ \frac{\mathrm{d}^3}{\mathrm{d}x^3} D &= (24x - 27 x^7) \cos x^3 - 81 x^4 \sin x^3 \xrightarrow{x \rightarrow 0} 0 \text{,} \\ \frac{\mathrm{d}^4}{\mathrm{d}x^4} D &= (24 - 432 x^6) \cos x^3 + (-396 x^3 + 81 x^9) \sin x^3 \xrightarrow{x \rightarrow 0} 24 \text{.} \end{align*} So the limit is $\frac{14}{24} = \frac{7}{12}$.