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.

Could anyone explain to me how to calculate the limit $$\lim_{x \to 0} \frac{1}{\sqrt{x^3}} - \frac1{\sin x}$$

share|improve this question

3 Answers 3

up vote 5 down vote accepted

First note that the limit only makes sense for $x \to 0^+$, as the function is otherwise undefined.

As it is posted, we have the indeterminate form $\infty - \infty$. So it makes sense to find the common denominator and add the two fractions.

Write $$\lim_{x\to 0} \left(\frac 1{\sqrt{x^3}} - \frac 1{\sin x}\right) = \lim_{x\to 0} \frac{\sin x - \sqrt{x^3}}{\sqrt{x^3}{\sin x}}$$

And then perform L'Hospital.

(Note: it will probably be easier to work with if you write $\sqrt{x^3} = x^{3/2}$.)

share|improve this answer
    
thanks, i tried to find solution, but there's no two-sided limit. Do i need to find two for 0+ and 0-? –  aine Jul 26 at 14:46
    
If we approach $0$ only from the right, then the limit is $\infty$. Approaching from the left is not defined in the reals. (Indeed, the function is not defined on the reals for $x<0$. –  amWhy Jul 26 at 14:55
    
thans again) now i understand) –  aine Jul 26 at 15:29
    
You're welcome, aine! –  amWhy Jul 26 at 15:35
    
Thanks, friend! –  amWhy Jul 27 at 15:28

We solve the problem without using L'Hospital's Rule, by making precise the intuition that as $x$ approaches $0$ from the right, the function $\frac{1}{x^{3/2}}$ blows up faster than $\frac{1}{\sin x}$.

(We have to look at the limit from the right, since our function is not defined for negative $x$.)

Since the limit as $x$ approaches $0$ of $\frac{\sin x}{x}$ is $1$, we have $\sin x\gt \frac{x}{2}$ if $x$ is positive and near enough to $0$. And it does not have to be really near.

Now suppose that $x\lt \frac{1}{16}$. Then $\sin x\gt \frac{x}{2}\gt 2x^{3/2}$.

It follows that if $0\lt x\lt \frac{1}{64}$ then $$\frac{1}{x^{3/2}}-\sin x\gt \frac{1}{x^{3/2}}-\frac{1}{2x^{3/2}}=\frac{1}{2x^{3/2}}.$$

But $\frac{1}{2x^{3/2}}\to\infty$ as $x\to 0^+$. So we conclude that $$\lim_{x\to 0^+} \left(\frac{1}{x^{3/2}}-\frac{1}{\sin x}\right)$$ does not exist, or if you prefer, that it is $\infty$.

share|improve this answer
    
great, thank you for detailed explanation. –  aine Jul 26 at 15:30
    
You are welcome. The L'Hospital Rule calculation gets somewhat unpleasant, though it is nicer if we replace $x$ by $t^2$. A series approach would also be natural. –  André Nicolas Jul 26 at 15:33
    
@AndréNicolas Shouldn't it be the case that $$\lim_{x \to 0^+}f(x)=\lim_{x \to 0^-} f(x)$$ in order for the limit to exist at $x=0$(even if it is infinite)? As amWhy has pointed out, $f(x)$ is undefined (in $\mathbb{R}$) for negative $x$, so $$\lim_{x \to 0^-} f(x)$$ does not exist. –  alexqwx Jul 26 at 15:41
    
The answer I wrote points out (second paragraph) that we can only consider limits from the right. I showed that the limit from the right does not exist, or alternately is infinite. Certainly the plain limit does not exist, there is no need to do any calculation to see that. –  André Nicolas Jul 26 at 15:45
    
But you said in your last sentence that the limit does exist and is $\infty$. –  alexqwx Jul 26 at 15:47

Hint

Add the two fractions, simplify, and use hopital

share|improve this answer

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.