4
$\begingroup$

Find the limit as $x$ approaches $0$ of

$\dfrac{e^{x^2} - \cos x}{\sin^2 x}$

What I tried is as $x$ approaches $0$, $e^{x^2}$ tends to $1$ and so the numerator tends to $1-\cos x$ and after doing some trigonometric simplifications I got the answer as $\frac 12$.

Is this right? Any help would be appreciated.

$\endgroup$
  • 1
    $\begingroup$ Your answer is incorrect. The numerator also tends to 0 because $\cos(0) = 1$ and $e^0=1$. $\endgroup$ – Jolien Apr 10 '15 at 8:42
  • $\begingroup$ There was $\frac{x}{2}$ before the edit, now there's $x^2$. Anyway, there's nothing tricky here, it's a simple application of the l'Hospital rule. $\endgroup$ – peter.petrov Apr 10 '15 at 8:44
  • $\begingroup$ As far as I read it from the beginning it was always $\;x^2\;$ , but perhaps the OP himself change it almost immediately after posted. $\endgroup$ – Timbuc Apr 10 '15 at 9:28
3
$\begingroup$

Using l'Hospital:

$$\lim_{x\to 0}\frac{e^{x^2}-\cos x}{\sin^2 x}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{2xe^{x^2}+\sin x}{2\sin x\cos x}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{2e^{x^2}+4x^2e^{x^2}+\cos x}{2\cos^2x-2\sin^2x}=\frac{2+0+1}{2}=\frac32$$

$\endgroup$
  • $\begingroup$ So far the only non-Taylor answer here. Hooray! $\endgroup$ – user198044 Apr 10 '15 at 8:49
  • 1
    $\begingroup$ @Jack It may be cultural. In France, I was mostly trained with Taylor series (or "développements limités"), and L'Hospital's rule was rather disregarded. I see Taylor series as more systematic and more powerful. $\endgroup$ – Jean-Claude Arbaut Apr 10 '15 at 8:51
  • 1
    $\begingroup$ @Jean-ClaudeArbaut I studied both methods in Israel, yet I have to say that I consider l'Hospital as the fastest, easiest and, in general, best method when appliable. Of course, sometimes doing some limits without l'H helps to develop certain skills (trigonometric identities, algebraic manipulation), etc. $\endgroup$ – Timbuc Apr 10 '15 at 9:10
  • $\begingroup$ @Jean-ClaudeArbaut Taylor before LHR? What about those 0^0 things? $\endgroup$ – user198044 Apr 10 '15 at 10:24
  • 1
    $\begingroup$ @jack Usually not a problem with $a^b=\exp(b \log a)$ and either Taylor, or basic limits such as $x^\alpha \log^\beta x$. When I was a student I have actually never used L'Hospital's rule. Not that it's that bad, but I prefered other tools, and it was taught that way anyway. $\endgroup$ – Jean-Claude Arbaut Apr 10 '15 at 10:29
3
$\begingroup$

HINT:

$$\frac{e^{x^2}-\cos x}{\sin^2x}=\cdots=\frac{\dfrac{e^{x^2}-1}{x^2}}{\left(\dfrac{\sin x}x\right)^2}+\frac{1-\cos x}{1-\cos^2x}$$

Use $\lim_{h\to0}\dfrac{e^h-1}h=1=\lim_{u\to0}\dfrac{\sin u}u$ for the first part

and as $x\to0,\cos x\to1\implies\cos x-1\ne0$ for the second

$\endgroup$
  • 1
    $\begingroup$ @Jean-ClaudeArbaut, Please revert the wrong rectification $\endgroup$ – lab bhattacharjee Apr 10 '15 at 8:59
  • $\begingroup$ Well, I don't see your point then. The fraction with cosines has a removable singularity, and it disappears when you cancel the common factor. $\endgroup$ – Jean-Claude Arbaut Apr 10 '15 at 9:00
  • $\begingroup$ @Jean-ClaudeArbaut, As $1-\cos x\ne0$ it can be cancelled safely $\endgroup$ – lab bhattacharjee Apr 10 '15 at 9:01
  • $\begingroup$ Ah, ok. Actually, it's not completely true: when you compute a limit as $x\rightarrow a$, you have also to consider the case $x=a$ (but see below). Here it cancels safely not because the denominator is not zero, but because the numerator is also zero, and there is a common factor, leaving only $1+\cos x$. But it's not wrong either: there is a known difference between french "limite" and english "limit" (which is usually "limite épointée" in french). And here the difference is important. $\endgroup$ – Jean-Claude Arbaut Apr 10 '15 at 9:01
  • $\begingroup$ @Jean-ClaudeArbaut I have to disagree with your comment in the first line: when considering limits as $\;x\to a\;$ we must disregard as principle the case $\;x=a\;$ and, in fact, it is "our right and duty" to assume $\;x\;$ gets arbitraritly close to $\;a\;$ yet it never actually reaches it. Many texts use this: Purcell et al., Strang, Stewart, Spivak, etc. Some even stress this by requiring $\;0<|x-a|<\delta\;$ in the definition. $\endgroup$ – Timbuc Apr 10 '15 at 9:26
1
$\begingroup$

Use Taylor expansion. As $x\rightarrow0$, you have

$$e^{x^2}=1+x^2+O(x^4)$$ $$\cos x=1-\frac12x^2+O(x^4)$$ $$\sin^2x=x^2+O(x^4)$$

Thus

$$\frac{e^{x^2}-\cos x}{\sin^2 x}=\frac{1+x^2-1+\frac12x^2+O(x^4)}{x^2+O(x^4)}=\frac{\frac32+O(x^2)}{1+O(x^2)}\underset{x\rightarrow 0}{\longrightarrow} \frac32$$

$\endgroup$
1
$\begingroup$

As Jean-Claude Arbaut showed, Taylor expansions are extremely useful for this kind of problems.

But, using them with a few more terms, you also see how is the limit approached $$A=\frac{e^{x^2}-\cos x}{\sin^2 x}=\frac{\Big(1+x^2+\frac{x^4}{2}+O\left(x^6\right)\Big)-\Big(1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)\Big)}{x^2-\frac{x^4}{3}+O\left(x^6\right)}$$ $$A\approx\frac{\frac{3 x^2}{2}+\frac{11 x^4}{24}+O\left(x^6\right)}{x^2-\frac{x^4}{3}+O\left(x^6\right)}=\frac{\frac{3 }{2}+\frac{11 x^2}{24}}{1-\frac{x^2}{3}}$$ Performing the long division, we have $$A\approx \frac{3}{2}+\frac{23 }{24}x^2$$ If you plot on the same graph the function and the above approximation, you will probably be amazed to see how close to each other are the two curves for $-\frac 12\leq x \leq \frac 12$.

$\endgroup$
0
$\begingroup$

You can use Taylor series as

$$\dfrac{e^{x^2} - \cos x}{\sin^2 x} \sim_{x\sim 0} \dfrac{(1+x^2) - (1-x^2/2)}{x^2} =\frac{3}{2}. $$

$\endgroup$

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.