Evaluate: lim x→0 (7^x − 1) / (2^x − 1)
Understand the Problem
The question is asking to evaluate the limit of the expression as x approaches 0. The expression contains exponential functions in both the numerator and denominator, and we need to determine the value of this limit.
Answer
$$ \frac{\ln(7)}{\ln(2)} $$
Answer for screen readers
The final answer is: $$ \frac{\ln(7)}{\ln(2)} $$
Steps to Solve
- Identify the limit expression
We need to evaluate the limit of the expression as $x$ approaches 0: $$ \lim_{x \to 0} \frac{7^x - 1}{2^x - 1} $$
- Apply the exponential limit property
Recall that $\lim_{x \to 0} a^x - 1 = \ln(a) \cdot x$ as $x$ approaches 0. Thus, we can rewrite the expression:
- For $7^x - 1 \approx \ln(7) \cdot x$ as $x \to 0$
- For $2^x - 1 \approx \ln(2) \cdot x$ as $x \to 0$
- Rewrite the limit using these approximations
Substituting these approximations into our limit gives: $$ \lim_{x \to 0} \frac{\ln(7) \cdot x}{\ln(2) \cdot x} $$
- Simplify the expression
Notice that $x$ cancels out: $$ \lim_{x \to 0} \frac{\ln(7)}{\ln(2)} $$
- Evaluate the limit
The limit simplifies directly to: $$ \frac{\ln(7)}{\ln(2)} $$
The final answer is: $$ \frac{\ln(7)}{\ln(2)} $$
More Information
This result can be interpreted as the ratio of the natural logarithms of 7 and 2, which gives us the logarithmic base change factor between these two numbers.
Tips
- A common mistake is to assume that $7^x$ and $2^x$ behave linearly as $x$ approaches 0, without using the proper limit properties.
- Misusing L'Hôpital's rule before confirming that the limit is in an indeterminate form (like 0/0).
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