Limits without using L'Hospital's rule, $\lim _{x\to -1}\frac{ e ^{x^2-1}-1}{\ln(-x)}$

Question

I would like to calculate these limits without using L'Hospital:

$$\lim _{x\to -1}\frac{ e ^{x^2-1}-1}{\ln(-x)}$$

$$\lim _{x\to 0}\frac{ ({\ln(1-x^2)}+1)^{(1/4)} -1}{x\sin2x}$$

In the second I tried to multiply by $$ {\ln(1-x^2)}+1)^{(1/4)} +1$$ but I think it worth nothing because I need to "vanish" the 0 in the denominator.

Any hints, please?

Answer 1

A Path

For the first one, you might want to recall these two fundamental limits: $$\frac{e^{\alpha(x)}-1}{\alpha(x)}\to 1,$$ and $$\frac{\log (1+\alpha(x))}{\alpha(x)}\to 1$$ when $$\alpha(x) \to 0.$$ Therefore \begin{eqnarray} \mathcal L_1 &=& \lim_{x\to -1} \frac{e^{x^2-1}-1}{\log (-x)}=\\ &=&\lim_{x\to -1}\underbrace{\frac{e^{x^2-1}-1}{x^2-1}}_{\to 1}\cdot \underbrace{\frac{-x-1}{\log (1+(-x-1))}}_{\to 1}\cdot \frac{x^2-1}{-x-1}. \end{eqnarray} Can you proceed from here?

For the second problem, I would use the other fundamental limits $$\frac{(1+\alpha(x))^k-1}{\alpha(x)}\to k$$ and, of course $$\frac{\sin \alpha(x)}{\alpha(x)} \to 1,$$ again for $\alpha(x) \to 0$. Can you do it this way?

Answer 2

Set $\frac{1}{4}=n$:

$$\lim _{x\to 0}\frac{\left(\ln\left(1-x^{2}\right)+1\right)^{\frac{1}{4}}-1}{x\sin\left(2x\right)}$$$$=\lim _{x\to 0}\frac{\color{blue}{\left(\ln\left(1-x^{2}\right)+1\right)^{\frac{1}{4}}-1}}{1}\cdot\frac{{1}}{x\sin\left(2x\right)}$$$$=\lim _{x\to 0}\frac{1}{\color{blue}{\sum_{k=0}^{n-1}\left(\ln\left(1-x^{2}\right)+1\right)^{\left(\frac{k}{n}\right)}}}\cdot\frac{\color{blue}{\ln\left(1-x^{2}\right)}}{-x^{2}}\cdot\frac{-\color{green}{2}x}{\sin\left(2x\right)}\cdot\frac{1}{\color{green}{2}}=-\frac{1}{8}$$

The other one has been answered by another users.

Answer 3

Hints:

1. $$\frac{e^{x^2-1}-1}{\log(-x)}=\frac{e^{x^2-1}-1}{x^2-1}\frac{(x+1)(x-1)}{\log(-x)}=\frac{e^{x^2-1}-1}{x^2-1}\frac{e^{\log(-x)}-1}{\log(-x)}(1-x)\to1\cdot1\cdot2.$$

2.

Notice that $$\sqrt[4]{a+1}-1=\frac{(a+1)-1}{(a+1)^{3/4}+(a+1)^{2/4}+(a+1)^{1/4}+1}.$$

If $a$ tends to $0$, the denominator tends to $4$.

Then

$$\frac{\log(1-x^2)}{x\sin 2x}=\frac12\frac{\log(1-x^2)}{x^2}\frac{2x}{\sin 2x}=-\frac12\left(\frac{e^{\log(1-x^2)}-1}{\log(1-x^2)}\right)^{-1}\frac{2x}{\sin 2x}\to-\frac12\cdot1^{-1}\cdot1.$$

Make sure that in the expressions of the form $\frac{e^t-1}t$, $t$ does tend to $0$.