Q.Discuss the continuity of the following function at the point(s) or on the interval indicated against it: $f(x) = \dfrac{|x+1|}{2x^2+x-1}$, for $x \ne -1$, $= 0$ for $x=-1$, at $x=-1$.

$f(x)=\dfrac{|x+1|}{2x^2+x-1}$ for $x\ne-1$, $f(-1)=0$.

Factor the denominator: $2x^2+x-1=(2x-1)(x+1)$.

For $x>-1$ (so $x+1>0$): $|x+1|=x+1$, so $f(x)=\dfrac{x+1}{(2x-1)(x+1)}=\dfrac{1}{2x-1}$. $\displaystyle\lim_{x\to-1^+} f(x)=\frac{1}{2(-1)-1}=\frac{1}{-3}=-\frac13$.

For $x<-1$ (so $x+1<0$): $|x+1|=-(x+1)$, so $f(x)=\dfrac{-(x+1)}{(2x-1)(x+1)}=\dfrac{-1}{2x-1}$. $\displaystyle\lim_{x\to-1^-} f(x)=\frac{-1}{2(-1)-1}=\frac{-1}{-3}=\frac13$. …

Split into two cases based on the sign of $x+1$ (since $|x+1|$ behaves differently on each side of $x=-1$), simplify the resulting rationa …

Treating $|x+1|/(x+1)$ as always equal to $1$ regardless of the sign of $x+1$ — it equals $+1$ only when $x>-1$ and $-1$ when $x<-1$, which …