Does L' Hospital's rule pay-off at all in calculating: $$\displaystyle\lim_{x\to 0}\frac{\sqrt{\cos2x}\cdot e^{2x^2}-1}{\ln{(1+2x)\cdot\ln{(1+2\arcsin{x})}}}$$

I posted this question on Quora not expecting the answers on finding the limit itself, but to eliminate all the bad options. Since we haven't formally gone through L'Hospital's rule, I tried avoiding it because it seemed like a cliche. I took the commutativity between continuous functions & limits, as well. Since $x\to 0$, I thought I could maybe substitute $x$ by $\frac{1}{y}$ when $y\to\infty$. However, that wasn't helpful either. I wasn't sure which terms I can replace by another function as they aren't the same. For example $\sqrt{\cos{2x}}\;\&\;e^{2x^2}$.

I examined an answer by Paramanand Singh (prefer his approach):

Solve without L'Hopital's rule: $\lim_{x\to0}\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}$

At some point, I realized my attempts of manipulating are all inadequate.

My question: how to choose the function substitutes & is there any other approach involving pure algebraic fractions manipulation? And any other suggestions/opinions on whether L'Hospital pays-off?

We haven't gone through so much, which is worrying & now we have to work all on our own (I don't complain, that's great & motivating in most cases, but sometimes is rather difficult without a proper literature) Steppan Konoplev's solution:

Let $f(x),g(x)$ be the numerator, denominator respectively. Since $$\cos x \approx 1 - \frac{x^2}{2}, \sqrt{1+x} \approx 1+ \frac{x}{2}, e^x \approx 1+x,$$ we have:$$f(x) \approx \sqrt{1-2x^2}(1+2x^2) - 1 \approx (1-x^2)(1+2x^2)-1 = x^2-2x^4\;\text{around}\;x=0$$

My note: $e^{2x^2}\approx1+2x^2\;$?

On the other hand: $\arcsin x \approx x, \ln(1+x) \approx x\;$so we have:$$g(x) \approx \ln(1+2x)^2 \approx 4x^2\;\text{around}\;x=0$$ Thus: $$\frac{f(x)}{g(x)} = \frac{x^2+O(x^4)}{4x^2+O(x^4)}\to\frac{1}{4}\;\text{as}\;x \to 0.$$

I also read there:

The simplified form of the numerator is easy, but the denominator is messy if you write out all the terms in detail. It does seem that it would be easier to use the first few terms of power series expansions.

  • $\begingroup$ Why don't you apply the technique mentioned in my linked answer? $\endgroup$ – Paramanand Singh Jan 9 at 7:39
  • $\begingroup$ The denominator here is rather easy. Just rewrite it as $$2x\cdot\dfrac{\log(1+2x)}{2x}\cdot 2x\cdot\frac{\arcsin x} {x} \cdot\frac {\log(1+2\arcsin x)} {2\arcsin x} $$ and the denominator is then safely replaced by $4x^2$. $\endgroup$ – Paramanand Singh Jan 9 at 7:44
  • $\begingroup$ For numerator add and subtract $e^{2x^2}$. Or if you wish you may add or subtract $\sqrt{\cos 2x}$. $\endgroup$ – Paramanand Singh Jan 9 at 7:46
  • $\begingroup$ @ParamanandSingh, First, I apologise for misspelling your name. I was trying to aply that to the whole expression not seeing the obvious pattern $\displaystyle\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$ $\endgroup$ – Invisible Jan 9 at 8:31
  • 1
    $\begingroup$ Don't worry i fixed the spelling. $\endgroup$ – Paramanand Singh Jan 9 at 8:35

My preferred approach is to use a set of well kwown limits combined with algebraic manipulation. This is a simple approach which requires familiarity with laws of limits. Techniques like L'Hospital's Rule and Taylor series are powerful but require some care (especially in case of L'Hospital's Rule one must first ensure that prerequisites for the rule are met).

On the other hand the usage of well known limits requires some deeper understanding. Thus consider for example the limit $$\lim_{x\to 0}\dfrac{\log(1+x)}{x}=1$$ One should notice that the argument of log function ie $(1+x)$ tends to $1$ and the denominator tends to $0$. Thus whenever you see an expression like $$\log\text{(something)} $$ where "something" tends to $1$ you have to express it as $$\frac{\log(\text{something})} {\text{something} - 1}\cdot(\text{something} - 1)$$ and see if this helps you or not.

This kind of understanding is needed for all the well known limits. For example can you figure out what to do with $\sqrt{\cos 2x}$ in numerator (see current question) given the well known limit $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}?$$ Ask yourself the same question about $e^{2x^2}$ in numerator given the well known limit $$\lim_{x\to 0}\frac{e^x-1}{x}=0$$ Another rule of thumb is that use L'Hospital's Rule only when required differentiation is damn easy (no calculations, just memory) and resulting expression is simpler. For example you can apply it on $$\lim_{x\to 0}\frac{x-\sin x} {x^3}$$ to get $$\lim_{x\to 0}\frac{1-\cos x} {3x^2}$$ If you need to use product rule of differentiation for applying L'Hospital's Rule then you are better off not applying it.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your patience & understanding! For: $\lim_{x\to 0}\frac{1-\sqrt{\cos(2x)}}{x^2}=\lim_{x\to 0}\frac{1-\sqrt{\cos(x)^2-\sin(x)^2}}{x}=\lim_{x\to 0}\frac{1-\cos{x}}{x}?$ I hope I'll improve by praticing! Thank you once again for the beautiful explanation! $\endgroup$ – Invisible Jan 9 at 10:59
  • $\begingroup$ @ParamanandSingh, May I ask, in this question:math.stackexchange.com/questions/3134961/…, is there a better option than MacLaurin series? I didn't use $L'Hospital$, but tried to put $\frac{1}{x^{2018}}$ into the exponent so as to apply the definition of $e$ in the $1^{\text{st}}$ term & $\frac{1}{x}$ into the exponent in the $2^{\text{nd}}$ term $2018\;\text{times repeatedly}$ because of the $2018-\text{th}$ power of $\ln$. The problem was $1'\text{s}$ annullated & there was still $x$ in the denomenator, so, indeterminate form. $\endgroup$ – Invisible Jan 9 at 22:18
  • 1
    $\begingroup$ @VerkhovtsevaKatya: I have added an answer to question which you mention in your comment. However I don't fully understand your technique mentioned in comment. Can you post a separate question describing your approach and ways to fix it? Note that questions which contain asker's approach are encouraged here. $\endgroup$ – Paramanand Singh Jan 10 at 2:32
  • $\begingroup$ @ParamanandSingh, thank you for your kind words, but, unfortunately, I cannot post more than a certain number of questions per a few weeks, so I commented on your answer to the mentioned question. $\endgroup$ – Invisible Jan 10 at 6:16

$$\lim_{x\to0}\dfrac{\sqrt{\cos2x}\cdot e^{2x^2}-1}{\ln(1+2x)\cdot\ln(1+2\arcsin x)}$$

$$=\dfrac14\cdot\lim_{x\to0}\dfrac{\cos2x\cdot e^{4x^2}-1}{x^2}\cdot\dfrac1{\lim_{x\to0}\dfrac{\ln(1+2x)}{2x}}\cdot\dfrac1{\lim_{x\to0}\dfrac{\ln(1+2\arcsin x)}{2\arcsin x}}\cdot\dfrac1{\lim_{x\to0}\dfrac{\arcsin x} x}\cdot\lim_{x\to0}\dfrac1{\sqrt{\cos2x}\cdot e^{2x^2}+1}$$

Only the first limit deserves further treatment

$$\lim_{x\to0}\dfrac{\cos2x\cdot e^{4x^2}-1}{x^2}=\lim_{x\to0}\dfrac{(1-2\sin^2x)(e^{4x^2}-1)}{x^2}=\lim_{x\to0}\dfrac{e^{4x^2}-1}{x^2}-2\lim_{x\to0}\dfrac{\sin^2x}{x^2}(e^{4x^2}-1)=4$$

| cite | improve this answer | |

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.