Answer
$-1$
Work Step by Step
Let us consider $f(x)=\lim\limits_{x \to 1}\dfrac{5x^2-7x+2}{x^2-1}$
Now, $x=1.001 \implies f(x)= -0.999000; \\x=1.01 \implies f(x)= -0.990000 \\x=1.1 \implies f(x)=-0.900000$
From the computed data it has been seen that as $x$ approaches $-1$ from left and right, then $f(x)$ approaches $6$, so we have:
$ LHL =\lim\limits_{x \to 1^{-}}f(x)=-1$ and $ RHL =\lim\limits_{x \to 1^{+}}f(x)=-1$
Since, $ LHL =RHL$
Therefore, $\lim\limits_{x \to 1}f(x)=-1$
