Question

If $f(x) = \frac{4^{x - \pi} + 4^{\pi - x} - 2}{(x - \pi)^2}$, for $x \ne \pi$, is continuous at $x = \pi$, then $f(\pi)$ is

• A. $2(\log 2)^2$
• B. $4(\log 2)^2$
• C. $(\log 2)^2$
• D. $8(\log 2)^2$

Hint:

Any limit in the form $\lim_{x \to 0} \frac{a^x + a^{-x} - 2}{x^2}$ perfectly wraps up into $(\log a)^2$. Recognizing that $a^x + a^{-x} - 2$ is the expansion of $(a^{x/2} - a^{-x/2})^2$ saves a lot of time!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
For a function to be continuous at a point, the limit of the function as it approaches that point must equal the function's value at that point. Thus, $f(\pi) = \lim_{x \to \pi} f(x)$.

Step 2: Key Formula or Approach:
We will use substitution to simplify the limit and apply the standard logarithmic limit identity:
$$\lim_{t \to 0} \frac{a^t - 1}{t} = \log a$$

Step 3: Detailed Explanation:
The limit we need to evaluate is:
$$L = \lim_{x \to \pi} \frac{4^{x - \pi} + 4^{-(x - \pi)} - 2}{(x - \pi)^2}$$ To make it easier, let $t = x - \pi$. As $x \to \pi$, $t \to 0$.
The limit becomes:
$$L = \lim_{t \to 0} \frac{4^t + 4^{-t} - 2}{t^2}$$ Multiply the numerator and denominator by $4^t$ to eliminate the negative exponent:
$$L = \lim_{t \to 0} \frac{4^{2t} + 1 - 2(4^t)}{t^2 \cdot 4^t}$$ Notice that the numerator is a perfect square: $4^{2t} - 2(4^t) + 1 = (4^t - 1)^2$.
$$L = \lim_{t \to 0} \frac{(4^t - 1)^2}{t^2 \cdot 4^t}$$ Separate the terms to apply the standard limit:
$$L = \lim_{t \to 0} \left( \frac{4^t - 1}{t} \right)^2 \cdot \lim_{t \to 0} \frac{1}{4^t}$$ Using the identity $\lim_{t \to 0} \frac{4^t - 1}{t} = \log 4$:
$$L = (\log 4)^2 \cdot \frac{1}{4^0}$$ $$L = (\log 4)^2 \cdot 1$$ We can simplify this further using logarithm properties ($\log 4 = \log 2^2 = 2\log 2$):
$$L = (2\log 2)^2 = 4(\log 2)^2$$ Since the function is continuous, $f(\pi) = L = 4(\log 2)^2$.

Step 4: Final Answer:
The value of $f(\pi)$ is $4(\log 2)^2$, matching option (B).