0
$\begingroup$

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

My steps so far:

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

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

  • 2
    Your first line after "My steps so far" is wrong – 2017-02-19
  • 1
    You have used the equation $\log(a + b) = \log(a)\log b$ instead of the correct one $\log(ab) = \log a + \log b$. Perhaps the equation got switched somehow when you were trying to solve the problem. – 2017-02-20

2 Answers 2

3

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

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

  • 0
    My book's brief answer states that the value of the limit is $1/e.$ How would I get to that value? – 2017-02-19
  • 0
    @MaxLi Recall that $$1=\lim_{x\to0}\frac{\ln(1+x)}x$$ – 2017-02-19
  • 0
    Can you provide a source? Thanks. – 2017-02-19
  • 1
    @MaxLi Before tackling this problem, your book should surely have a source of the following: $$\frac d{dx}\ln(x)\bigg|_{x=1}=\lim_{h\to0}\frac{\ln(1+h)}h=\lim_{h\to1}\frac{\ln(h)}{h-1}=1$$ – 2017-02-19
1

$\dfrac{\ln(e+h)-1}h\;$ 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 $x=e$.