Question

\(\lim _{x\rightarrow 1}\left[\frac{2^{2x-2}-2^{x}+1}{\sin ^{2}(x-1)}\right]=\)

• A. \(1\)
• B. \((\log 2)^{2}\)
• C. \(\log 2\)
• D. \(2\,\log 2\)

Hint:

Put x - 1 equal to t, factor the numerator as a perfect square and use standard limits.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Let \(t=x-1\). As \(x\to1\), \(t\to0\).

Step 2: Rewrite the Numerator
\(2^{2x-2}-2^{x}+1=2^{2t}-2\cdot2^{t}+1=(2^{t}-1)^{2}\)
because \(2^{x}=2\cdot2^{t}\).

Step 3: Expression
The limit becomes \(\lim_{t\to0}\dfrac{(2^{t}-1)^{2}}{\sin^{2}t}\).

Step 4: Use Standard Limits
Divide numerator and denominator by \(t^{2}\).
\(\lim\dfrac{2^{t}-1}{t}=\log2\) and \(\lim\dfrac{\sin t}{t}=1\).
So the limit is \((\log2)^{2}\), option (B).

Final Answer:
The limit equals the square of log 2. \[ \boxed{\text{(B) }(\log 2)^{2}} \]