Find $\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$ Find $$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$$
My work so far:
$$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}=\frac{\ln3-\ln5}{\ln4-\ln10}$$
Is correct?
Add:
I used $a^x\sim 1+x\ln a$ for $x\rightarrow 0$
 A: L'Hospital's rule for $\frac{0}{0}$ limits works well in this case. An alternate is to use 
$$a^{x} = e^{x \, \ln(a)} = 1 + \ln(a) \, x + \frac{\ln^{2}(a)}{2!} \, x^{2} + \mathcal{O}(x^{3})$$
which yields
\begin{align}
\frac{a^{x} - b^{x}}{c^{x} - d^{x}} &= \frac{(\ln(a) - \ln(b)) \, x + \frac{1}{2} \, (\ln^{2}(a) - \ln^{2}(b)) \, x^{2} + \mathcal{O}(x^{3})}{(\ln(c) - \ln(d)) \, x + \frac{1}{2} \, (\ln^{2}(c) - \ln^{2}(d)) \, x^{2} + \mathcal{O}(x^{3})} \\
&= \frac{\ln(\frac{a}{b}) + \frac{1}{2} \, \ln(\frac{a}{b}) \, \ln(a b) \, x + \mathcal{O}(x^{2})}{\ln(\frac{c}{d}) + \frac{1}{2} \, \ln(\frac{c}{d}) \, \ln(c d) \, x + \mathcal{O}(x^{2})}
\end{align} 
Taking the limit as $x \to 0$ yields
$$\lim_{x \to 0}  \frac{a^{x} - b^{x}}{c^{x} - d^{x}} = \frac{\ln(\frac{a}{b})}{\ln(\frac{c}{d})} = \frac{\ln (a) - \ln(b)}{\ln(c) - \ln(d)}.$$
By L'Hospital's rule:
\begin{align}
\lim_{x \to 0}  \frac{a^{x} - b^{x}}{c^{x} - d^{x}} &= \lim_{x \to 0}  \frac{e^{x \, \ln(a)} - e^{x \, \ln(b)}}{e^{x \, \ln(c)} - e^{x \, \ln(d)}} \\
&= \lim_{x \to 0} \frac{\ln(a) \, a^{x} - \ln(b) \, b^{x}}{\ln(c) \, c^{x} - \ln(d) \, d^{x}} \\
&= \frac{\ln (a) - \ln(b)}{\ln(c) - \ln(d)}.
\end{align}
A: The expression equals
$$\frac{(3^x-3^0) - (5^x-5^0)}{(4^x - 4^0)-(10^x-10^0)}.$$
Divide top and bottom by $x=x-0.$ Then by definition of the derivative, the desired limit equals
$$ \frac{(3^x)'(0) - (5^x)'(0)}{(4^x)'(0)- (10^x)'(0)}= \frac{\ln 3 - \ln 5}{\ln 4 - \ln 10}.$$
A: HINT: you can write your result in the form $$\frac{\ln\left(\frac{5}{3}\right)}{\ln\left(\frac{5}{2}\right)}$$
