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).