Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 66

Estimate the value of the $\displaystyle \lim_{x \to 0} \frac{5^x - 4^x}{3^x - 2^x}$ by using its graph. Then use L'Hospital's Rule to find the exact value.




Based from the graph the value of the limit as $x$ approaches 0 is $y \approx 0.52$. Now by using L'Hospital's Rule...

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0} \frac{5^x - 4^x}{3^x - 2^x} &= \lim_{x \to 0} \frac{5^x (\ln 5) - 4^x (\ln 4)}{3^x(\ln 3)- 2^x(\ln 2)}\\
\\
&= \frac{5^0 (\ln 5) - 4^0 (\ln 4)}{3^0 (\ln 3) - 2^0 (\ln 2)} = \frac{\ln 5 - \ln 4}{\ln 3 - \ln 2}
\end{aligned}
\end{equation}
$


By applying the Laws of Logarithm, we have...
$\displaystyle = \frac{\ln \frac{5}{4}}{\ln \frac{3}{2}}$

Suppose that $f(x) = 2x \sin x$ and $g(x) = \sec x - 1$