Find lim as x to 0 (a^x-b^x)/(c^x-d^x), where a,b,c,d > 0
Find $\displaystyle\lim_{x\to0}\frac{a^{x}-b^{x}}{c^{x}-d^{x}}$, where $a,b,c,d\gt 0$ and $c\neq d$.
Explanation of Method
Four exponentials, and only their logarithms remain.
Step 1: The form
Each of the four powers equals $1$ at $x = 0$, so numerator and denominator both vanish: the form is $\dfrac00$.
Step 2: Apply l'Hôpital's rule
\[ \lim_{x\to0}\frac{a^{x}-b^{x}}{c^{x}-d^{x}} = \lim_{x\to0}\frac{a^{x}\ln a-b^{x}\ln b}{c^{x}\ln c-d^{x}\ln d}. \]
The condition $c\neq d$ guarantees $\ln c-\ln d\neq0$, so the new denominator does not vanish at $x = 0$ and the substitution is legitimate.
Step 3: Substitute $x = 0$
\[ \frac{\ln a-\ln b}{\ln c-\ln d} = \frac{\ln\dfrac{a}{b}}{\ln\dfrac{c}{d}}. \]
Step 4: Verification
\[a,b,c,d = 3,2,5,4:\ \frac{\ln1.5}{\ln1.25} = \frac{0.4054651}{0.2231436} = 1.8170595\]
\[x = 0.01:\ 1.8061548; \qquad x = 0.001:\ 1.8159660 \;\;\checkmark\]
\[\text{Taking } c = e,\ d = 1 \text{ reduces this to Problem 385's answer } \ln\tfrac{a}{b} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\lim_{x\to0}\frac{a^{x}-b^{x}}{c^{x}-d^{x}} = \frac{\ln\dfrac{a}{b}}{\ln\dfrac{c}{d}}\,}\]
Remark
Remark. Both differences are equivalent to linear functions at the origin: $a^{x}-b^{x}\sim x\ln\tfrac{a}{b}$ and $c^{x}-d^{x}\sim x\ln\tfrac{c}{d}$. The factor $x$ cancels, which is why only the ratio of logarithms survives. Note also that the answer is unchanged if all four bases are raised to the same power, since that multiplies both logarithms alike.