Question

\( \lim_{x \to 2} \frac{x^2 + 2^2 - 5}{2^{x-2} - 2} \) is equal to ____

• A. \( \frac{2}{\ln 2} \)
• B. \( \frac{4}{\ln 2} \)
• C. \( 4\ln 2 \)
• D. \( 2\ln 2 \)

Hint:

Always check if limit is \( 0/0 \) or \( \infty/\infty \) before applying L'Hospital's Rule.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Substitute \( x = 2 \) into the expression:
Numerator: \( 2^2 + 2^0 - 5 = 4 + 1 - 5 = 0 \).
Denominator: \( 2^{-1} - 2^{-1} = 0 \).
Since it is in \( \frac{0}{0} \) form, we can use L'Hôpital's Rule or algebraic substitution.
Step 2: Key Formula or Approach:
Let \( 2^{x/2} = t \). As \( x \to 2, t \to 2 \).
The expression becomes:
\[ \lim_{t \to 2} \frac{t^2 + 4/t^2 - 5}{1/t - 2/t^2} \]
Step 3: Detailed Explanation:
Simplifying the fractions:
\[ \frac{\frac{t^4 + 4 - 5t^2}{t^2}}{\frac{t - 2}{t^2}} = \frac{t^4 - 5t^2 + 4}{t - 2} \]
Factoring the numerator:
\[ t^4 - 5t^2 + 4 = (t^2 - 1)(t^2 - 4) = (t-1)(t+1)(t-2)(t+2) \]
Substituting back:
\[ \lim_{t \to 2} \frac{(t-1)(t+1)(t-2)(t+2)}{t - 2} = \lim_{t \to 2} (t-1)(t+1)(t+2) \]
Evaluating at \( t = 2 \):
\[ (2-1)(2+1)(2+2) = 1 \times 3 \times 4 = 12 \]
Step 4: Final Answer:
The value of the limit is 12.