Question

The value of \(f(0)\) so that \(\frac{-e^x+2^x}{x}\) may be continuous at \(x=0\) is

• A. \(\log\left(\frac{1}{2}\right)\)
• B. 0
• C. 4
• D. \(-1+\log 2\)

Hint:

Use expansions: \(e^x=1+x+\dots\) and \(a^x=1+x\ln a+\dots\) for limits at \(x\to 0\).

Answer 1

The Correct Option is D

Step 1: Condition for continuity at \(x=0\).
To make the function continuous at \(x=0\), we must define:
\[ f(0)=\lim_{x\to 0}\frac{-e^x+2^x}{x} \]
Step 2: Rewrite expression.
\[ \lim_{x\to 0}\frac{2^x-e^x}{x} \]
Step 3: Use standard limit forms.
We know:
\[ \lim_{x\to 0}\frac{a^x-1}{x}=\ln a \quad \text{and} \quad \lim_{x\to 0}\frac{e^x-1}{x}=1 \]
Step 4: Split into two limits.
\[ \frac{2^x-e^x}{x} =\frac{(2^x-1)-(e^x-1)}{x} =\frac{2^x-1}{x}-\frac{e^x-1}{x} \]
Step 5: Apply limits.
\[ \lim_{x\to 0}\frac{2^x-1}{x}=\ln 2 \] \[ \lim_{x\to 0}\frac{e^x-1}{x}=1 \]
So:
\[ f(0)=\ln 2 - 1 \]
Final Answer:
\[ \boxed{-1+\log 2} \]