Question:
Evaluate \(\lim_{x\rightarrow2}(x^{2}-3x+3)^{\frac{1}{x^{2}-4}}\).
Evaluate \(\lim_{x\rightarrow2}(x^{2}-3x+3)^{\frac{1}{x^{2}-4}}\).
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Whenever you encounter \(1^\infty\), take logarithms first and convert the problem into a \(0/0\) or \(\infty/\infty\) form.
Updated On: Jun 9, 2026
- \(e^{1/2}\)
- 0
- \(e^{1/4}\)
- 1
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The Correct Option is C
Solution and Explanation
Concept:
As \(x\to2\), \[ x^2-3x+3 \to 1 \] and \[ \frac1{x^2-4}\to\infty \] Hence the limit is of the indeterminate form \(1^\infty\).
Step 1: Let the limit be \(L\). \[ L= \lim_{x\to2} \left(x^2-3x+3\right)^{\frac1{x^2-4}} \] Take logarithm: \[ \ln L = \lim_{x\to2} \frac{\ln(x^2-3x+3)}{x^2-4} \] This is \(0/0\), so apply L'Hospital's Rule.
Step 2: Differentiate numerator and denominator. \[ \ln L = \lim_{x\to2} \frac{\frac{2x-3}{x^2-3x+3}}{2x} \] Substituting \(x=2\), \[ \ln L = \frac{\frac{1}{1}}{4} = \frac14 \]
Step 3: Find \(L\). \[ L=e^{1/4} \] \[ \boxed{e^{1/4}} \]
As \(x\to2\), \[ x^2-3x+3 \to 1 \] and \[ \frac1{x^2-4}\to\infty \] Hence the limit is of the indeterminate form \(1^\infty\).
Step 1: Let the limit be \(L\). \[ L= \lim_{x\to2} \left(x^2-3x+3\right)^{\frac1{x^2-4}} \] Take logarithm: \[ \ln L = \lim_{x\to2} \frac{\ln(x^2-3x+3)}{x^2-4} \] This is \(0/0\), so apply L'Hospital's Rule.
Step 2: Differentiate numerator and denominator. \[ \ln L = \lim_{x\to2} \frac{\frac{2x-3}{x^2-3x+3}}{2x} \] Substituting \(x=2\), \[ \ln L = \frac{\frac{1}{1}}{4} = \frac14 \]
Step 3: Find \(L\). \[ L=e^{1/4} \] \[ \boxed{e^{1/4}} \]
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