ISI2018-DCG-28

Question

Let $f(x)=e^{-\big( \frac{1}{x^2-3x+2} \big) };x\in \mathbb{R} \: \: \& x \notin \{1,2\}$. Let $a=\underset{n \to 1^+}{\lim}f(x)$ and $b=\underset{x \to 1^-}{\lim} f(x)$. Then

• A. $a=\infty, \: b=0$
• B. $a=0, \: b=\infty$
• C. $a=0, \: b=0$
• D. $a=\infty, \: b=\infty$

Comments

$\mathbf B$?

Answer 1

here given,

$f(x)=e^{-\frac{1}{x^{2}-3x+2}}$

=> $f(x)=e^{-\frac{1}{(x-1)(x-2)}}$

Now,

$a=\lim_{x->1^{+}}f(x)$

we can think it like we are coming from right side of 1 , means the value is slightly greater than 1 but not 1.

let us take the value as x=1.001,

now $f(1.001)=e^{-\frac{1}{(1.001-1)(1.001-2)}}$

                       = $f(1.001)=e^{1001.001}->\infty$ so as x->1, $f(x)->\infty$

$b=\lim_{x->1^{-}}f(x)$

 here can think it like we are coming from left side of 1, means the value is slightly lesser than 1 but not 1 .

let us take the value x=0.999

now $f(0.999)=e^{-\frac{1}{(0.999-1)(0.999-2)}}$

                      =$f(0.999)=e^{-999.00099}->0$

so correct answer is option (A).