Question

evaluate the limit using lhospitals rule \\(\lim_{x\to 0} \frac{10^x - 13^x}{x}\\) question help: video message instructor submit question jump to answer

Explanation:

Step1: Check the form of the limit

When \(x = 0\), \(\frac{10^{x}-13^{x}}{x}=\frac{10^{0}-13^{0}}{0}=\frac{1 - 1}{0}=\frac{0}{0}\), which is an indeterminate form. So, L'Hospital's rule can be applied. L'Hospital's rule states that if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\)

Step2: Differentiate the numerator and the denominator

The derivative of \(y = 10^{x}\) is \(y^{\prime}=10^{x}\ln(10)\) (using the formula \((a^{x})^{\prime}=a^{x}\ln(a)\)), the derivative of \(y = 13^{x}\) is \(y^{\prime}=13^{x}\ln(13)\), and the derivative of \(y=x\) is \(y^{\prime}=1\)

So, \(\lim_{x
ightarrow0}\frac{10^{x}-13^{x}}{x}=\lim_{x
ightarrow0}\frac{(10^{x}-13^{x})^{\prime}}{x^{\prime}}=\lim_{x
ightarrow0}(10^{x}\ln(10)-13^{x}\ln(13))\)

Step3: Evaluate the new limit

Substitute \(x = 0\) into \(10^{x}\ln(10)-13^{x}\ln(13)\)

When \(x = 0\), \(10^{0}\ln(10)-13^{0}\ln(13)=\ln(10)-\ln(13)\)

Using the property of logarithms \(\ln(a)-\ln(b)=\ln(\frac{a}{b})\), we have \(\ln(10)-\ln(13)=\ln(\frac{10}{13})\)

Answer:

\(\ln(\frac{10}{13})\)