Question

Evaluate \(\lim_{x\rightarrow2}(x^{2}-3x+3)^{\frac{1}{x^{2}-4}}\).

• A. \(e^{1/2}\)
• B. 0
• C. \(e^{1/4}\)
• D. 1

Hint:

Whenever you encounter \(1^\infty\), take logarithms first and convert the problem into a \(0/0\) or \(\infty/\infty\) form.

Answer 1

The Correct Option is C

Step 1: Spot the form.
As $x\to2$, the base $x^2-3x+3\to 4-6+3=1$ and the exponent $\dfrac{1}{x^2-4}\to\infty$. This is the $1^\infty$ form, so we use the standard trick.
Step 2: Use the exponential identity.
For $1^\infty$ limits, $\lim f^g=e^{\lim g(f-1)}$. Here $f-1=x^2-3x+2=(x-1)(x-2)$.
Step 3: Form the exponent.
\[ g(f-1)=\frac{(x-1)(x-2)}{x^2-4}=\frac{(x-1)(x-2)}{(x-2)(x+2)}. \]
Step 4: Cancel the common factor.
For $x\ne2$ this is $\dfrac{x-1}{x+2}$.
Step 5: Take the limit of the exponent.
\[ \lim_{x\to2}\frac{x-1}{x+2}=\frac{1}{4}. \]
Step 6: Exponentiate.
So the original limit is $e^{1/4}$.
\[ \boxed{e^{1/4}} \]