Why is $\lim_{x \rightarrow \infty} (1+\frac{1}{x})^x = e$?

[Zarlucicil]

This is a pretty basic limit question regarding the limit,

$$\lim_{x \rightarrow \infty} (1+\frac{1}{x})^x = e$$

Wolframalpha gives the following reasoning for this answer:

$$\lim_{x \rightarrow \infty} (1+\frac{1}{x})^x = e^{\lim_{x \rightarrow \infty} x\ln{(1+\frac{1}{x})}} = e^{\lim_{t \rightarrow 0} \frac{\ln{(1+t)}}{t}} = e^{\lim_{t \rightarrow 0} \frac{ \frac{d\ln{(1+t)}}{dt}}{\frac{d}{dt}t}}= e^{\lim_{t \rightarrow 0} \frac{1}{1+t}} = e^1$$

My question is, by the same reasoning, why is the following not true? (where log is log base 10)

$$\lim_{x \rightarrow \infty} (1+\frac{1}{x})^x = 10^{\lim_{x \rightarrow \infty} x\log{(1+\frac{1}{x})}} = 10^{\lim_{t \rightarrow 0} \frac{\log{(1+t)}}{t}} = 10^{\lim_{t \rightarrow 0} \frac{ \frac{d\log{(1+t)}}{dt}}{\frac{d}{dt}t}}= 10^{\lim_{t \rightarrow 0} \frac{1}{1+t}} = 10^1$$

Am I missing something??

 

[Zarlucicil]

Wooooow, hold on, I understand what I did wrong...

The derivative of $$\log{(1+t)}$$ is NOT $$\frac{1}{1+t}$$ but rather $$\frac{1}{(1+t)\ln{10}}$$

Sorry for wasting the time of those who've read this.