evaluate the limit using lhospitals rule\nlim...
evaluate the limit using lhospitals rule\nlim(x→0) (13^x - 5^x)/x
Answer
# Explanation:
## Step1: Check indeterminate form
When \(x = 0\), \(\frac{13^{x}-5^{x}}{x}=\frac{13^{0}-5^{0}}{0}=\frac{1 - 1}{0}=\frac{0}{0}\), so L'Hopital's rule can be applied.
## Step2: Differentiate numerator and denominator
The derivative of \(y = 13^{x}-5^{x}\) using the formula \((a^{x})^\prime=a^{x}\ln a\) is \(y^\prime=13^{x}\ln13 - 5^{x}\ln5\), and the derivative of \(y = x\) is \(y^\prime = 1\).
## Step3: Evaluate new limit
\(\lim_{x\rightarrow0}\frac{13^{x}\ln13 - 5^{x}\ln5}{1}\). Substitute \(x = 0\) into the expression: \(13^{0}\ln13-5^{0}\ln5=\ln13-\ln5\).
# Answer:
\(\ln13-\ln5\)