How to determine $\lim_{h \to 0} \frac{1- 9^h}{h}$

Question

How to determine $\lim_{h \to 0} \frac{1- 9^h}{h}$. Is it $3\log 3$?

Actually my question is a sub part of another question in which I have solved everything but this.

Answer 1

$\frac {d}{dx} e^x = \lim_\limits{h\to 0} \frac {e^{x+h} - e^x}{h} = e^x \lim_\limits{h\to 0} \frac {e^h-1}{h} = e^x\\ \lim_\limits{h\to 0} \frac {e^h-1}{h} = 1\\ \lim_\limits{h\to 0} -\frac {9^h-1}{h} = \lim_\limits{h\to 0} -\frac {e^{h\ln 9}-1}{h} =\lim_\limits{h\to 0} -\frac {(e^{h\ln 9}-1)\ln 9}{h\ln 9}.$

Let $u = \ln 9.$

$\lim_\limits{h\to 0} -\frac {\ln 9(e^{h\ln 9}-1)}{h\ln 9} = \lim_\limits{u\to 0} -\frac {\ln 9(e^{u}-1)}{u} = - \ln 9$

Answer 2

Note that we have :

$$\frac{1-9^h}{h} = \frac{e^0-e^{h\ln 9}}{h}$$

Hence :

$$\lim_{h \to 0} \frac{1-9^h}{h} = -(x \mapsto e^{x\ln 9})’(0)$$

Answer 3

Two things you should do to correct your answer:

1) Watch signs. Put in some small positive values of $h$ and you will see the limit is negative.

2) We do not have $\log (mn)=n \log m$. Rather $\log(m^n)=n\log m$.

Answer 4

$$\lim_{h\to0}\frac{1-9^h}{h}=\lim_{h\to0}\frac{1-e^{h\ln9}}{h} \\ =\lim_{h\to0}\dfrac{1-\left(1+\dfrac{h\ln(9}{1!}+\dfrac{(h\ln(9))^2}{2!}+\cdots\right)}{h} \\ =\lim_{h\to0}-\left(\dfrac{\ln(9}{1!}+\dfrac{h(\ln(9))^2}{2!}+\cdots\right) \\ =-\ln(9)$$