Discussion on choosing the optimal method in finding:$\lim_{x\to 0}\frac{\sqrt{\cos2x}\cdot e^{2x^2}-1}{\ln{(1+2x)\cdot\ln{(1+2\arcsin{x})}}}$

Question

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.

Answer 1

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.

Answer 2

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

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

Only the first limit deserves further treatment

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