Determine $$\lim_{h \to 0}\frac{\ln(e+h)-1}{h}$$
My steps so far:
$$=\lim_{h \to 0}\frac{\ln e \ln h-1}{h}$$
$$=\lim_{h \to 0}\frac{\ln h -1}{h}$$
Asked
Active
Viewed 2,866 times
0
-
2Your first line after "My steps so far" is wrong – Feb 19 '17 at 22:47
-
1You 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. – Paramanand Singh Feb 20 '17 at 01:26
2 Answers
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
$$\lim_{h\to0}\frac{\ln(e+h)-1}h=\frac1e\lim_{x\to0}\frac{\ln(1+x)}x$$
Simply Beautiful Art
- 74,685
-
My book's brief answer states that the value of the limit is $1/e.$ How would I get to that value? – Jack Pan Feb 19 '17 at 22:54
-
-
-
1@MaxLi Before tackling this problem, your book should surely have a source of the following: $$\frac d{dx}\ln(x)\bigg|{x=1}=\lim{h\to0}\frac{\ln(1+h)}h=\lim_{h\to1}\frac{\ln(h)}{h-1}=1$$ – Simply Beautiful Art Feb 19 '17 at 23:08
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 $x=e$.
Bernard
- 175,478