Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Calculate the limit:

$$\lim_{x \to 0} \frac{\cos x - e^{x^2}}{x \sin x}$$

So I have gone back and forward with this question,

The problem is that the expansions I make from $\cos$ and $e$ are a bit problematic because of the grade of expansion. The restterm is having the same degree as the degree of $e$'s expansion.

Any kind of hint or guidance is appreciated :)

The answer is $-3/2$

share|improve this question
    
How far have you expanded $e^{x^2}$? –  Daniel Fischer Aug 24 at 21:05
    
I saw that my expansions were a bit too much and an small error too . –  Ara Aug 24 at 21:19

2 Answers 2

up vote 2 down vote accepted

Since $$\cos(x)=1-\frac{x^2}{2}+o(x^3),\qquad e^{x^2}=1+x^2+o(x^3),\quad x\sin x=x^2+o(x^3)$$ your limit is just: $$\lim_{x\to 0}\frac{(1-x^2/2)-(1+x^2)}{x^2}=-\frac{3}{2}.$$ You can state the same by applying De l'Hospital's rule twice.

share|improve this answer
    
Why is it so common to end on $o(3rd term)$? Does the term shrink down for most expressions by this point? –  Display Name Aug 25 at 1:28

The top has Maclaurin expansion $$\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)-\left(1+\frac{x^2}{1!}+\frac{x^4}{2!}+\cdots\right).$$

This simplifies to $-\frac{3}{2}x^2-\frac{11}{24}x^4+\cdots$.

The bottom has expansion $$x^2-\frac{x^4}{3!}+\cdots.$$

Divide top and bottom by $x^2$, and the limit will be clear.

share|improve this answer
    
Okey I see my error, I expanded it more than n = 2, if we call n the grade of expansion. (I don't know if you can call it that but I hope you understand) And also expanded the $e^{x^{2}}$ a bit wrong :/ –  Ara Aug 24 at 21:18
    
"Too many terms" is no problem, apart from the extra work involved. Expanding $e^{x^2}$ wrongly could mess up the conclusion. The best method, as you know, is to write down the expansion of $e^t$ and then replace $t$ everywhere by $x^2$. –  André Nicolas Aug 24 at 21:25
    
Ohh... thats correct what you are sayin. It was just the expansions. But one additional question... Do all the expansions have to be the same grade? Because just at the glance i think it has to be. –  Ara Aug 24 at 22:29
    
The problem you looked at is typical. Expand each term of the top. How far? Some of the terms may "cancel." We have to expand one step beyond that. So in our case the constant terms $1$ cancel, and so do the (invisible) terms in $x$. The first non-cancelling place for the expansion on top involves the $x^2$ terms. At the bottom, there is no cancellation, so the first term will be enough. –  André Nicolas Aug 24 at 23:12
    
Okey, got it! @André Nicolas So we can have different terms we look at the numerator and denominator separately. –  Ara Aug 25 at 11:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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