(d) $$ \lim _{x ightarrow 2}\left\{\begin{array}{ll}1-x & x \leq 2 \ 3 x^{2}+1 & x2\end{array} ight. $$

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(d) $$ \lim _{x ightarrow 2}\left\{\begin{array}{ll}1-x & x \leq 2 \ 3 x^{2}+1 & x>2\end{array}ight. $$

UpStudy AI · Accepted Answer

To evaluate the limit

$$
\lim_{x \rightarrow 2} f(x) = \lim_{x \rightarrow 2} \left\{
\begin{array}{ll}
1 - x & \text{if } x \leq 2 \
3x^2 + 1 & \text{if } x > 2
\end{array}
\right.
$$

we need to consider the left-hand limit and the right-hand limit separately.

### Left-Hand Limit ($$x \rightarrow 2^-$$)

For values of $$x$$ approaching 2 from the left ($$x \leq 2$$), the function is defined as $$f(x) = 1 - x$$. Therefore,

$$
\lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^-} (1 - x) = 1 - 2 = -1
$$

### Right-Hand Limit ($$x \rightarrow 2^+$$)

For values of $$x$$ approaching 2 from the right ($$x > 2$$), the function is defined as $$f(x) = 3x^2 + 1$$. Therefore,

$$
\lim_{x \rightarrow 2^+} f(x) = \lim_{x \rightarrow 2^+} (3x^2 + 1) = 3(2)^2 + 1 = 12 + 1 = 13
$$

### Conclusion

Since the left-hand limit and the right-hand limit are not equal ($$-1 \neq 13$$), the overall limit does not exist.

$$
\boxed{\text{The limit does not exist because the left and right limits are unequal.}}
$$
The limit does not exist because the left-hand limit is -1 and the right-hand limit is 13, which are not equal.

(d)

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