Jorden Pace

Answered

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 $\underset{x\to 0}{lim}\frac{{5}^{x}\mathrm{ln}5-{3}^{x}\mathrm{ln}3}{1}=\underset{x\to 0}{lim}(\mathrm{ln}5-\mathrm{ln}3)=\mathrm{ln}\frac{5}{3}$

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 $\underset{x\to 0}{lim}\frac{{5}^{x}\mathrm{ln}5-{3}^{x}\mathrm{ln}3}{1}=\underset{x\to 0}{lim}(\mathrm{ln}5-\mathrm{ln}3)$. I am not sure how the derivative becomes what it does, nor do I understand the next step. I do understand $\underset{x\to 0}{lim}(\mathrm{ln}5-\mathrm{ln}3)=\mathrm{ln}\frac{5}{3}$, however.

Thanks!

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 $\underset{x\to 0}{lim}\frac{{5}^{x}\mathrm{ln}5-{3}^{x}\mathrm{ln}3}{1}=\underset{x\to 0}{lim}(\mathrm{ln}5-\mathrm{ln}3)=\mathrm{ln}\frac{5}{3}$

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 $\underset{x\to 0}{lim}\frac{{5}^{x}\mathrm{ln}5-{3}^{x}\mathrm{ln}3}{1}=\underset{x\to 0}{lim}(\mathrm{ln}5-\mathrm{ln}3)$. I am not sure how the derivative becomes what it does, nor do I understand the next step. I do understand $\underset{x\to 0}{lim}(\mathrm{ln}5-\mathrm{ln}3)=\mathrm{ln}\frac{5}{3}$, however.

Thanks!

Answer & Explanation

furniranizq

Expert

2022-07-05Added 20 answers

${5}^{x}={e}^{x\mathrm{ln}5}$

Now use $\frac{d}{dx}{e}^{x}={e}^{x}$ and the chain rule:

$\frac{d}{dx}{5}^{x}=\frac{d}{dx}{e}^{x\mathrm{ln}5}=\mathrm{ln}(5)\cdot {e}^{x\mathrm{ln}5}=\mathrm{ln}(5)\cdot {5}^{x}$

Now use $\frac{d}{dx}{e}^{x}={e}^{x}$ and the chain rule:

$\frac{d}{dx}{5}^{x}=\frac{d}{dx}{e}^{x\mathrm{ln}5}=\mathrm{ln}(5)\cdot {e}^{x\mathrm{ln}5}=\mathrm{ln}(5)\cdot {5}^{x}$

pouzdrotf

Expert

2022-07-06Added 4 answers

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

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