 Jorden Pace

2022-07-04

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:
$\underset{x\to 0}{lim}\frac{{5}^{x}-{3}^{x}}{x}$
Solution:
Determining that this function has indeterminate form $0/0$, we apply L’Hospital’s rule.
Applying L.H. rule, we get $\underset{x\to 0}{lim}\frac{{5}^{x}\mathrm{ln}5-{3}^{x}\mathrm{ln}3}{1}=\underset{x\to 0}{lim}\left(\mathrm{ln}5-\mathrm{ln}3\right)=\mathrm{ln}\frac{5}{3}$
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 $\underset{x\to 0}{lim}\frac{{5}^{x}\mathrm{ln}5-{3}^{x}\mathrm{ln}3}{1}=\underset{x\to 0}{lim}\left(\mathrm{ln}5-\mathrm{ln}3\right)$. I am not sure how the derivative becomes what it does, nor do I understand the next step. I do understand $\underset{x\to 0}{lim}\left(\mathrm{ln}5-\mathrm{ln}3\right)=\mathrm{ln}\frac{5}{3}$, however.
Thanks! furniranizq

Expert

${5}^{x}={e}^{x\mathrm{ln}5}$
Now use $\frac{d}{dx}{e}^{x}={e}^{x}$ and the chain rule:
$\frac{d}{dx}{5}^{x}=\frac{d}{dx}{e}^{x\mathrm{ln}5}=\mathrm{ln}\left(5\right)\cdot {e}^{x\mathrm{ln}5}=\mathrm{ln}\left(5\right)\cdot {5}^{x}$ pouzdrotf

Expert

You're just letting $x\to 0$. Since ${5}^{0}={3}^{0}=1$, those factors disappear.