Question

Another One, getting boring

Answer 1

\(\lim\limits_{x \to 0} ~ ~ \dfrac{e^{2x}-1}{\ln(x+1)}\)

Answer 2

Using l'hopitals rule gives:

\[\lim_{x \rightarrow 0}\frac{ 2e^{2x} }{ \frac{ 1 }{ x+1 }}\]

this gives us a limit of 2 when we plug in x

I can't think of a way to do this w/o using l'hopes

Answer 3

Owned!!

Alternative [better 😏] solution is the Taylor/Maclaurin expansion: everytime, if you can squeeze it in, separately on numerator and denominator. \(f(x) \approx f(0) + x f'(0) + \cdots \)

This one is a peach, though:

\(\lim\limits_{x \to 0^+} x \ln x\)

PS

That one can be Hôpitalised algebraically and thusly trivialised. But your natural aversion to L'H suggests you are well schooled or discerning. L'H is deliberately not taught in many well-versed learning communities, eg the Indians and the French. So there is a better solution than the L'H black-box.

There always is.