1. ## Another limit (known)

Prove that $\lim_{x\to0}\frac{\ln(1+x)}x=1$

Of course, without L'Hôpital's Rule

2. We can use the series expansion of ln(1+x)

$ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-............$

Dividing by x gives:

$1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\frac{x^{4}}{5}-..........$

Of course, we can see if x approaches 0, we are left with 1.

3. Okay

Now, without series?

4. Hello, Krizalid!

Recall the definitions of $e$.

. . $e \;\;=\;\;\lim_{x\to0}(1 + x)^{\frac{1}{x}} \;=\;\lim_{z\to\infty}\left(1 + \frac{1}{z}\right)^z$

Prove that: . $\lim_{x\to0}\frac{\ln(1+x)}x=1$

Of course, without L'Hôpital's Rule

$\text{We have: }\;\frac{\ln(1+x)}{x} \;\;=\;\;\frac{1}{x}\!\cdot\!\ln(1+x) \;\;=\;\;\ln(1+x)^{\frac{1}{x}}$

$\text{Then: }\;\lim_{x\to0}\left[\ln(1+x)^{\frac{1}{x}}\right] \;\;=\;\;\ln\underbrace{\left[\lim_{x\to0}(1+x)^{\frac{1}{x}} \right] }_{\text{This is }e} \;\;=\;\;\ln e \;\;=\;\;1$

5. Yeah

I had exactly the same proof

Cheers,
K.