Detailed Solution: Limits
Evaluate the following limit: $$ \lim_{x \to 0} \frac{a^{2x} - 1}{x} $$
Solution:
This limit takes the indeterminate form of $\frac{0}{0}$ when $x = 0$. We can evaluate it using the standard exponential limit formula: $\lim_{t \to 0} \frac{k^t - 1}{t} = \ln(k)$. Here are two common ways to solve it.
Method 1: Using Exponent Rules
We can rewrite the term $a^{2x}$ using the power rule for exponents, $(a^m)^n = a^{mn}$. This gives us $(a^2)^x$. Substituting this back into the limit expression:
$$ \lim_{x \to 0} \frac{(a^2)^x - 1}{x} $$
Now, applying the standard limit formula where our constant base is $a^2$, we get:
$$ \ln(a^2) $$
Using the logarithm power property $\ln(x^y) = y \ln(x)$, this simplifies to:
$$ 2 \ln(a) $$
Method 2: Algebraic Substitution
Alternatively, we can manipulate the denominator to perfectly match the exponent $2x$ by multiplying both the numerator and the denominator by $2$:
$$ \lim_{x \to 0} \left( \frac{a^{2x} - 1}{x} \times \frac{2}{2} \right) = \lim_{x \to 0} 2 \cdot \frac{a^{2x} - 1}{2x} $$
Since $x \to 0$ implies that $2x \to 0$, we can substitute $t = 2x$ and directly apply the standard formula $\lim_{t \to 0} \frac{a^t - 1}{t} = \ln(a)$:
$$ 2 \cdot \ln(a) $$
Final Answer:
$$ 2\ln(a) $$