I couldn't solve $\lim\limits_{h \to 0}\frac{e^{(x+h)\log a}-e^{x\log a}}{h}=1$

Question

I am so stuck here. I have no idea where do I start. Could anyone help me out with this:

$$\lim\limits_{h \to 0}\frac{e^{(x+h)\log (a)}-e^{x\log (a)}}{h}$$

Answer 1

Can you make a good choice of $f(x)$ ?

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]

Answer 2

Hint: What is $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\quad ?$$Now choose $f(x)$ appropriately and you'll see immediately what to do :)

Answer 3

An idea: Think about the definition of $f'(x)$ by using the limit. and then assume $f(x)=a^x$

Answer 4

This is the derivative of $a^x$ which equals $a^xln(a)$. Divide both sides by $ln(a)$, take the logarithm on both sides and you're almost done. By the way, it is $\frac{-ln(ln(a))}{ln(a)}$.

Answer 5

$$\lim\limits_{h \to 0}\frac{e^{(x+h)\log (a)}-e^{x\log (a)}}{h}$$

$$=e^{x\log a}\lim\limits_{h \to 0}\left(\frac{e^{h\log a}-1}h\right)$$

$$=e^{x\log a}\log a\lim\limits_{h \to 0}\left(\frac{e^{h\log a}-1}{h\log a}\right)\text { as } h\to0\implies h\log a\to 0\text{ for finite } a>0 $$

$$=a^x\log a \text { as } e^{m\log a}=(e^{\log_ea})^m=a^m$$

Answer 6

According to your header question, $$\frac{d}{dx} a^x = a^x \ln(a) = 1$$

I dont know if you meant to put that "$=1$" in there, but you did.

There is no solution. There is no singular value for $a$ that makes this equality true for all values of $x$.