4. Let f(x)= ) x^2+1,&xleqslant 1 x+2,&xgt 1f(x) does not exist.

To show that the limit of the piecewise function $f(x)$ as $x$ approaches 1 does not exist, we need to check the left-hand limit and the right-hand limit separately.<br /><br />The left-hand limit is the limit of $f(x)$ as $x$ approaches 1 from the left. In this case, $f(x) = x^2 + 1$ for $x \leq 1$. Evaluating the left-hand limit, we have:<br /><br />$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 1) = 1^2 + 1 = 2$<br /><br />The right-hand limit is the limit of $f(x)$ as $x$ approaches 1 from the right. In this case, $f(x) = x + 2$ for $x > 1$. Evaluating the right-hand limit, we have:<br /><br />$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + 2) = 1 + 2 = 3$<br /><br />Since the left-hand limit and the right-hand limit are not equal, the limit of $f(x)$ as $x$ approaches 1 does not exist.