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$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}$$ In the book I am reading, the limit evaluated in this way: $$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}\times \frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}=\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{x^2+x}\times\left(\sqrt{x^2+x+1}+1\right)$$

Then it used equivalence and wrote:

$$\lim_{x\to 0^+}\frac{2x\left(\frac12x^2\right)}{x^2+x}\times\left(\sqrt{x^2+x+1}\right)=\lim_{x\to 0^+}\frac{x^3}{x(x+1)}.\left(\sqrt{x^2+x+1}\right)=0$$

I wonder why we should do all these calculation? Is it possible to use L'hopital rule and get $\frac0{\tfrac12}=0$ ?

Etemon
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    Yes. If you check that the L'H Rule hypotheses hold. [I agree that the first step is a waster of time, essentially the solution given is a Taylor Thm one and we should just plunge in.] – ancient mathematician Nov 01 '20 at 14:20
  • @ancientmathematician what about the first step. multiply to the fraction $\frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}$ . is it what we do generally in evaluating limits and when we have square root in the denominator and want to rationalize it ? – Etemon Nov 01 '20 at 14:25
  • But there's no need here, by Taylor I got numerator $x^3+HT$ and denominator $\frac{1}{2} (x+x^2)+HT$. – ancient mathematician Nov 01 '20 at 15:14

2 Answers2

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Short answer:

$$\tan x\sim x$$

and

$$\sqrt{x^2+x+1}-1\sim\frac x2.$$

Hence the factor $1-\cos x$ makes the limit tend to zero.

  • This is a nice discussion for an expert user but I exhort the asker to do not proceed in this way to solve limits, till they don't acquire the necessary experience to manage properly limit calculation by asymptotic equivalence. – user Nov 01 '20 at 14:35
  • @user I agree with you. I remember that you mentioned it to me before. but in this problem it seems very good. – Etemon Nov 01 '20 at 14:36
  • @soheil I mean, with the time and experience it is the standard way to proceed by a sight to evaluate the limit but really it can leads to wrong results if we use this way without the necessary attention. For example with $\frac{1- \cos x}{x}$ we could wrongly state that limit is zero since $\cos x \sim 1$ or with $\frac{x- \sin x}{x^2}$ also. Pay attention with it. – user Nov 01 '20 at 14:40
  • @soheil: just remember that you may multiply/divide the asymptotic approximations, but not add/subract them. $\cos x\sim 1$ does no imply $\cos x-1\sim 0$. Knowing this, you are safe. –  Nov 01 '20 at 14:49
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Yes since the expression is in the form $\frac 0 0$ we are allowed to use l'Hospital to obtain

$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}=\lim_{x\to 0^+}\frac{2 \sec^2 x - 2 \cos x}{\frac{2 x + 1}{2 \sqrt{x^2 + x + 1}}}=\frac{0}{\frac12}=0$$

Note that often use l'Hospital's rule is esplicitely not allowed because when we use it blindly we can't really understand what it is going to determine the limit. For these reason is good to solve limits in more than one way, using standard limits when possible.

In this case as an alternative we can proceed as follows

$$\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}=2\frac{\tan x}{x}\frac{1-\cos x}{x^2}\frac{x^3}{\sqrt{x^2+x+1}-1}\frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}$$

which allows to conclude by standard limits.

I also exhort not expert users to do not proceed blindly by asymptotic equivalence since it can leads to wrong results if we use this way without the necessary attention. For example as $x \to 0$ for $\frac{1- \cos x}{x^2}$ we could wrongly state that limit is zero since $\cos x \sim 1$ or with $\frac{x- \sin x}{x^3}$ also since $\sin x \sim x$.

user
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  • The OP knows that. –  Nov 01 '20 at 14:20
  • @YvesDaoust The OP is asking if it possible apply l'Hospital. I'll give also an alternative way. – user Nov 01 '20 at 14:21
  • I cannot speculate on answers that you will give. –  Nov 01 '20 at 14:35
  • @YvesDaoust I've indeed seprated the two parts. The first one is the answer to the main question "Is it valid to apply L'hopital rule to evaluate the limit?", the second part add a discussion about the general issue for the convenience to use different methods. – user Nov 01 '20 at 14:37
  • @soheil Yes I had in mind something else! Now fixed. For an expert user these are trivial but blindly using equivalence at the first order leads to a wrong result. We need to pay attention to that! – user Nov 01 '20 at 14:49
  • @user I got your point. thank you. it seems it need practice to recognize it. but what about L'hopital rule to check? it is always safe I think. – Etemon Nov 01 '20 at 14:54
  • @soheil In my opinion l'Hospital rule should'n never be used for all the limits we can solve by standard limits. Morover, I also think that Taylor's expansion is a better tools to solve limits because we get an ideea of what it is going on with the various term to determine the limit. There are very few cases for which we really need l'Hospital rule to solve and in any case it is better to use at least another way when possible. – user Nov 01 '20 at 15:01