Determine $\lim_{h \to 0}\frac{\ln(e+h)-1}{h}$

Question

Determine $$\lim_{h \to 0}\frac{\ln(e+h)-1}{h}$$

My steps so far:

$$=\lim_{h \to 0}\frac{\ln e \ln h-1}{h}$$

$$=\lim_{h \to 0}\frac{\ln h -1}{h}$$

Answer 1

Your first step is wrong, since

$$\ln(a+b)\ne\ln(a)\ln(b)$$

for all $a,b$.

Instead, notice that

$$\ln(e+h)-1=\underbrace{\ln(e+h)-\ln(e)=\ln\left(\frac{e+h}e\right)}_{\ln(a)-\ln(b)=\ln\left(\frac ab\right)}=\ln\left(1+\frac he\right)$$

If we then let $h=ex$, we end up with

$$\lim_{h\to0}\frac{\ln(e+h)-1}h=\frac1e\lim_{x\to0}\frac{\ln(1+x)}x$$

Answer 2

$\dfrac{\ln(e+h)-1}h\;$ is the rate of change of the function $\ln x$ from $x=e$ to $x=e+h$. So its limit is the derivative of $\ln x$ at the point $x=e$.