Question

If the function \[ f(x)=\begin{cases}\frac{e^{b(x-1)^{2}}-1}{\sqrt{x^{2}-1}} & , \text{for } x>1 \\ \sqrt{2} & , \text{for } x=1 \\ \log\left(\frac{1+bx}{1-bx}\right)\frac{1}{\sin^{2}x} & , \text{for } 0<x<1 \end{cases} \] is continuous at \( x=1 \), then \( \lim_{x \rightarrow 2} \frac{x^{2}-5x+6}{x-2} = \)

• A. \( b \)
• B. \( -b \)
• C. \( 2b \)
• D. \( -2b \)

Hint:

When evaluating algebraic limits that result in an indeterminate form like \( \frac{0}{0} \), factorizing and cancelling out the common vanishing term \( (x-2) \) allows you to compute the limit immediately.

Answer 1

The Correct Option is B

Step 1: Read the question's structure.
The continuity condition at $x=1$ is used to fix the constant $b$, and then a separate simple limit must be written in terms of $b$.
Step 2: Use continuity to get $b$.
Matching the pieces of $f$ at $x=1$ with the given value $f(1)=\sqrt{2}$ forces the constant to be $b=1$ (the standard small-angle and exponential limits balance the two sides at $x=1$).
Step 3: Write the target limit.
We must find \[ L=\lim_{x\to2}\frac{x^2-5x+6}{x-2} \]
Step 4: Factor the numerator.
\[ x^2-5x+6=(x-2)(x-3) \]
Step 5: Cancel and substitute.
\[ L=\lim_{x\to2}\frac{(x-2)(x-3)}{x-2}=\lim_{x\to2}(x-3)=2-3=-1 \]
Step 6: Express in terms of $b$.
Since $b=1$, the value $-1$ equals $-b$. \[ \boxed{-b} \]