I'm reading this text:
A few questions:
What's the importance of them going from $h$ to $x$ in the first line? What is the difference?
How did they go from $$\lim_{x \to 0} \frac{\ln(1+x) - \ln(1)}{x}$$ to $$\lim_{x \to 0} \left[\frac{1}{x} \cdot \ln(1+x)\right]$$
And then right before the blue 5 box... how did they go from: $$e^{\lim_{x \to 0} \ln(1+x)^{1/x}}$$ to $$\lim_{x \to 0} e^{\ln(1+x)^{1/x}}$$ How did they just pull out the limit sign?
2 Answers
$\begingroup$The x-h notations are equivalent.
Note that $\ln 1=0$ thus
$$\lim_{x \to 0} \frac{\ln(1+x) - \ln(1)}{x}=\lim_{x \to 0} \left(\frac{1}{x} \cdot \ln(1+x)\right)$$
Since the exponential function is continuos $$e^{\lim f(x)}\equiv \lim e^{f(x)}$$
$\endgroup$ 9 $\begingroup$There is no difference.
$\ln(1)=0$.
The exponential function is continuous.
