Does the L & # 39; Hospital rule pay off at all when 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 and didn't expect the answers to be answered when I found the limit myself, but to eliminate all bad options. We did not go through this formally *The rule of L & # 39; Hospital*I tried to avoid it because it seemed like a cliché.

I also examined the commutativity between continuous functions and boundaries.

Since $ x to $ 0I thought maybe I could replace it $ x $ by $ frac {1} {y} $ when $ y to infty $, But that wasn't helpful either.

I was not sure which terms I could replace with another function because they are not identical. For example $ sqrt { cos {2x}} ; & ; e ^ {2x ^ 2} $,

I have examined an answer from Paramarand Singh (prefer his approach):

Solve without the rule of L & # 39; Hopital: $ lim_ {x to0} frac { sqrt { cosh {(3x ^ 2)}} cdot e ^ {4x ^ 3} -1} {x ^ 2 tan (2x)} $

At some point I realized that my attempts at manipulation were all inadequate.

My question: How do you choose the functional substitutes? And any other suggestions / opinions about whether L & # 39; pays off?

We haven't been through so much, which is troubling, and now we all have to work alone (I'm not complaining, it's great and motivating in most cases, but sometimes it's pretty difficult without proper literature)

**Steppan Konoplev** Solution:

To let $ f (x), g (x) $ be the numerator or denominator. 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) approximately 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} approx. 1 + 2x ^ 2 ; $?

On the other hand: $ arcsin x approximately x, ln (1 + x) approximately x ; $so we

to have:$$ g (x) approximately ln (1 + 2x) ^ 2 approximately 4x ^ 2 ; text {around} ; x = 0 $$

Consequently: $$ 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 simple, but the denominator is

chaotic if you write out all the terms in detail. It seems so

would be easier to use the first terms of power expansions.