# Taking the logarithmic derivative of an exponential difference function after applying L'Hospital's Rule

Can somebody please explain the following application of L’Hospital’s Rule?

Find the limit:

$$\lim_{x\rightarrow 0} \frac{5^x-3^x}{x}$$

Solution:

Determining that this function has indeterminate form $0/0$, we apply L’Hospital’s rule.

Applying L.H. rule, we get $\lim_{x\to0} \frac{5^x\ln 5- 3^x\ln 3}{1} = \lim_{x\to 0}( \ln 5 - \ln 3) = \ln \frac53$.

The part that I am confused on is the application of the natural logarithm that occurs after the first application of L’Hospital’s Rule, namely $\lim_{x\to0} \frac{5^x\ln 5 - 3^x\ln 3}{1} = \lim_{x\to0}( \ln 5 - \ln 3)$. I am not sure how the derivative becomes what it does, nor do I understand the next step. I do understand $\lim_{x\to0} (\ln 5 - \ln 3) = \ln \frac53$, however.

Thanks!

$$5^x = e^{x\ln 5}$$
Now use $\frac{d}{dx} e^x = e^x$ and the chain rule:
$$\frac{d}{dx} 5^x = \frac{d}{dx} e^{x\ln 5} = \ln(5)\cdot e^{x\ln 5} = \ln(5)\cdot 5^x$$
You're just letting $x\to 0$. Since $5^0=3^0=1$, those factors disappear.