Find lim as x to 0 (a^x-b^x)/(tan x), where a > 0, b > 0
Find $\displaystyle\lim_{x\to0}\frac{a^{x}-b^{x}}{\tan x}$, where $a\gt 0$, $b\gt 0$.
Explanation of Method
Two exponentials separating at their own rates.
Step 1: The form
Since $a^{0} = b^{0} = 1$ and $\tan0 = 0$, both parts vanish: the form is $\dfrac00$.
Step 2: Apply l'Hôpital's rule
Using $\bigl(a^{x}\bigr)' = a^{x}\ln a$ and $\bigl(\tan x\bigr)' = \sec^{2}x$: \[\lim_{x\to0}\frac{a^{x}-b^{x}}{\tan x} = \lim_{x\to0}\frac{a^{x}\ln a - b^{x}\ln b}{\sec^{2}x} = \frac{\ln a-\ln b}{1}.\]
Step 3: Combine the logarithms
\[ \ln a-\ln b = \ln\frac{a}{b}. \]
Step 4: Verification
\[a = 3,\ b = 2:\ \ln\tfrac32 = 0.4054651\]
\[x = 0.001:\ 0.4058284; \qquad x = -0.001:\ 0.4051019 \;\;\checkmark\]
\[a = b:\ \text{the numerator is identically } 0,\ \text{and } \ln1 = 0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\lim_{x\to0}\frac{a^{x}-b^{x}}{\tan x} = \ln\frac{a}{b}\,}\]
Remark
Remark. Writing $a^{x} = e^{x\ln a}$ makes the mechanism plain: $a^{x}-b^{x} = x\bigl(\ln a-\ln b\bigr)+O(x^{2})$, and dividing by $\tan x\sim x$ leaves the bracket. The two numerical values above straddle $\ln\tfrac32$, one from each side, exactly as the $O(x^{2})$ term predicts.