Let $$ g(x)=\left\{\begin{array}{ll}x & ext { if } x

Question

Let $$ g(x)=\left\{\begin{array}{ll}x & 	ext { if } x<1 \ 3 & 	ext { if } x=1 \ 2-x^{2} & 	ext { if } 12\end{array}ight. $$ $$ \lim _{x ightarrow 2^{!}}[g(x)]=? $$

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To find the limit $$\lim _{x \rightarrow 2^{-}} g(x)$$, we need to determine the behavior of the function $$g(x)$$ as $$x$$ approaches 2 from the left (i.e., values of $$x$$ slightly less than 2).

The function $$g(x)$$ is defined piecewise as follows:

$$
g(x) =
\begin{cases}
x & \text{if } x < 1 \
3 & \text{if } x = 1 \
2 - x^{2} & \text{if } 1 < x \leq 2 \
x - 3 & \text{if } x > 2
\end{cases}
$$

For values of $$x$$ approaching 2 from the left ($$x \to 2^{-}$$), the relevant piece of the function is $$2 - x^{2}$$, since $$1 < x \leq 2$$.

Calculating the limit:

$$
\lim _{x \rightarrow 2^{-}} g(x) = \lim _{x \rightarrow 2^{-}} (2 - x^{2}) = 2 - (2)^2 = 2 - 4 = -2
$$

**Answer:** $$-2$$
The limit as $$x$$ approaches 2 from the left is $$-2$$.

Let

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