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I get confused between when to apply L'Hospital Rule and Taylor Series.
Is there any set of trigger points in the questions, that would be easier to solve with Taylor Series?

For Example, If the denominator is in terms of a large power of $x (>3)$, then L'Hospital Rule usually becomes complicated and is not advised.

Edit: Solve

$\lim _{x\to 0}\left(\left(\sin x\right)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right)$

The options given are: $0, 1, -1, \infty$

user
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Harshil Bhatt
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  • Do you have a specific example? You're generalising too much when there are a lot of limits you could simplify with L'Hopital's or Taylor series. – Toby Mak Dec 09 '18 at 05:29
  • @TobyMak, I meant the limits where L'Hospital Rule cannot be applied. I'll add an example in a few minutes. I just wanted to know if there are certain situations where we should use Taylor Expansions. – Harshil Bhatt Dec 09 '18 at 05:34
  • My point still holds. Generally speaking, if there are trigonometric functions such as $\sin x, \cos x, \ln x$, or exponential functions such as $e^x$ and $\ln x$, then the Taylor expansions will be easier. – Toby Mak Dec 09 '18 at 05:36
  • Have a look to this small nightmare : https://math.stackexchange.com/questions/925916/how-can-i-find-this-limit-involving-thrice-iterated-logarithm/925936#925936 – Claude Leibovici Dec 09 '18 at 06:30
  • You should first try manipulating the given expression to make use of standard limits. Most limit problems on this website don't need advanced tools like L'Hospital's Rule or Taylor. Even when these tools are needed the prior manipulation greatly simplifies the use of these tools. – Paramanand Singh Dec 09 '18 at 14:01
  • Your limit makes sense only when $x\to 0^{+}$ and the first term obviously tends to $0$. The second term tends to $1$ via the standard limits $x\log x\to 0$ and $(\sin x) /x\to 1$. – Paramanand Singh Dec 09 '18 at 14:04
  • @Paramanand: obviously? – TonyK Dec 11 '18 at 13:03
  • @TonyK: The first term is of the form $0^{\infty}$ which is not an indeterminate. A slightly more difficult example is $(1/2)^{1/x}\to 0$ as $x\to 0^{+}$. – Paramanand Singh Dec 12 '18 at 01:03
  • @ParamanandSingh: yes, OK. – TonyK Dec 12 '18 at 02:23

3 Answers3

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A general rule to follow is that we always should try at first to apply the simplest method for any limit. Therefore, when we face with an indeterminate form, I suggest at first to try with:

1. Algebraic manipulation, aimed to eliminate the source of that indetermination, e.g.

$$\lim_{x\to 1} \frac{x^2-3x+2}{x^2-1}=\lim_{x\to 1} \frac{(x-1)(x-2)}{(x-1)(x+1)}=\lim_{x\to 1} \frac{x-2}{x+1}=-\frac12$$

2. Standard limits, e.g.

$$x\to 0, \quad \frac{\sin x}x\to 1, \quad \frac{e^x-1}{x}\to 1, \quad x\log x \to 0,\quad \ldots$$

or related results derived from them, see for example: Are all limits solvable without L'Hôpital Rule or Series Expansion.

When we can't solve a limit in an indeterminate form by that basic tools, then we are allowed to use more advanced methods which involve derivatitive concepts and notably l'Hopital rule and Taylor's expansion.

In my opinion, with some special exception and when we are not explicitly requested/forced to use l'Hopital, we always should prefer Taylor's expansion not only because it is a more powerful and effective method but also because it allows us to really understand what is going on and which are the terms which really count in the expression whereas l'Hopital is a blind/black box method which doesn't add any contribution to our knowledge about limits.

Therefore if you know Taylor's expansion, in general, I suggest to proceed always by that.

Edit

The given example is one of the first category that is "solvable by elementary methods".

Notably we have that as $x \to 0^+$ (otherwise the expression in not defined)

$$\left(\sin x\right)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}=\left(\sin x\right)^{\frac{1}{x}}+\frac{1}{x^{\sin x}}\to 0 +1=1$$

Indeed we have that

  • $\left(\sin x\right)^{\frac{1}{x}}\to 0$ (it is in a not-indeterminate form $0^{\infty}$)

  • $x^{\sin x}=e^{\sin x \log x}\to e^0=1$

since by standard limits

$$\sin x \log x= \frac{\sin x}x\cdot x\log x\to 1 \cdot 0 =0$$

user
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  • But in an examination scenario, when time is of utmost importance, L'Hospital Rule would be the best way. – Harshil Bhatt Dec 11 '18 at 17:11
  • @HarshilBhatt Of course in some cases it can be useful to check the result for simple limit but, in generl, I suggest to do not learn limits by l'Hopital rule, it is really not helpful. In general, for limit not solveble by elementary methods, once we know Taylor's expansion, that's the preferable and more effective way to solve limits. – user Dec 11 '18 at 17:18
  • @HarshilBhatt Sorry I didn't take a close look to that but the given example can be of course solved by standard limit. I add something on that. – user Dec 11 '18 at 17:27
  • @HarshilBhatt: in a competitive exam scenario Taylor is far more effective provided you are comfortable with manipulation of series (like multiplying, dividing and composing them). – Paramanand Singh Dec 12 '18 at 01:05
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I should say that most of the time (not to say all the times) Taylor expansions are the keys.

Let me consider your example $$\lim _{x\to 0}\left(\left(\sin(x)\right)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin(x)}\right)$$

Consider the first term $$a=\left(\sin(x)\right)^{\frac{1}{x}} \implies \log(a)={\frac{1}{x}}\log\left(\sin(x)\right)$$ Now, by Taylor $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\log\left(\sin(x)\right)=\log (x)-\frac{x^2}{6}+O\left(x^4\right)$$ $$\log(a)=\frac{\log (x)}{x}-\frac{x}{6}+O\left(x^3\right)$$ So $\log(a)\to -\infty$ which means thet $a\to 0$.

Now, the second term (using equivalent is faster here) $$b=\left(\frac{1}{x}\right)^{\sin(x)}\implies \log(b)=-\sin(x) \log(x)\sim -x \log(x)\to 0^+$$ which means that $b\to 1^+$ and so $a+b$.

Claude Leibovici
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When you see, as $f\to0$, the following functions, it is often a good way to use Taylor series rather than L'Hospital rule.

  • $\log(1+f) = f+o(f)$
  • $\sin f = f+o(f)$
  • $\cos f = 1+ o(1)$
  • $e^f = 1+o(1)$
  • $\tan f = f+o(f)$
  • $\arctan f = f+o(f)$

(More terms of the series are sometimes necessary).

Lorenzo B.
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  • What you are indicating are the first order expansion which can be proved solely by standard limits and do no requres derivative concept. Moreover when we deal with limits I always suggest to include the remainder term, that is $\log(1+x)=x+o(x)$ etc., the asympthotic notation is really dangerous. – user Dec 12 '18 at 07:33
  • Moreover your suggestion, "when we see...then" is not a good guide in general. The first try should always be done with simpler method (algebraic manipulation, standard limits). – user Dec 12 '18 at 07:35
  • @gimusi I would say using Taylor series is quite a simple method, once you understand the meaning of it. I just wrote down the first order expansions to give an example; as I said, more terms of the series are sometimes necessary. – Lorenzo B. Dec 12 '18 at 14:37
  • Yes I agree with you, Taylor's series is of course the most powerfull and effective tool for limits but, for an educational purpouse, in my opinion at first we should always try to use elementary methods whenever possible. For the order of the expansions my main concern was for the remainder which I suggest to include always when dealing with limits. Bye – user Dec 12 '18 at 14:47
  • It is always better to include remainders, but yesterday I didn't have time to include them. Bye – Lorenzo B. Dec 12 '18 at 14:58
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    That's was exactly my main point. Thanks, Bye – user Dec 12 '18 at 15:02