It Might Not Exist

If we want to compute the two-sided limits , we first need to compute both their one-sided limits. In this case, if $\displaystyle \lim_{x\to2^-} f(g(x)) = \lim_{x\to2^+} f(g(x)) = L$ for some finite $L$ , then the limit in question exists and is equal to $L$ .

Let us solve the left hand limit first:

$\begin{aligned} \displaystyle \lim_{x\to2^-} f(g(x)) &=& \displaystyle \lim_{x\to 2^-} f(3-x) , \qquad \because \text{ if } x\to2^- , \text{ then } 3-x \to 1^+ \\ &=& \displaystyle \lim_{x\to 2^-} 3(3-x) - 1 \\ &=& 3(3-2) - 1 = 2 \end{aligned}$

Similarly, for the right hand limit:

$\begin{aligned} \displaystyle \lim_{x\to2^+} f(g(x)) &=& \displaystyle \lim_{x\to 2^+} f(2x-3) , \qquad \because \text{ if } x\to2^+ , \text{ then } 2x-3\to 1^+ \\ &=& \displaystyle \lim_{x\to 2^+} 3(2x-3) - 1 \\ &=& 3(2\cdot2 - 3) - 1 = 2 \end{aligned}$

Since $\displaystyle \lim_{x\to2^-} f(g(x)) = \lim_{x\to2^+} f(g(x))$ is indeed fulfilled, then the limit in question, $\displaystyle \lim_{x\to2} f(g(x))$ , exists and is equal to $\boxed2$ .

When $x=2$ , then $f(x) =3x-1$ and $g(x) =2x-3$ . Therefore,

$\begin{aligned} \lim_{x \to 2} f(g(x)) & = \lim_{x \to 2} 3(2x-3)-1 \\ & =3(4-3)-1=\boxed{2} \end{aligned}$