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

Question

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

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To find $$\lim _{x \rightarrow 1} g(x)$$ for the given piecewise function:

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

we need to evaluate the left-hand limit ($$x \to 1^-$$) and the right-hand limit ($$x \to 1^+$$).

1. **Left-Hand Limit ($$x \to 1^-$$):**
   
   For $$x < 1$$, $$g(x) = x$$.

$$
   \lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} x = 1
   $$

2. **Right-Hand Limit ($$x \to 1^+$$):**
   
   For $$1 < x \leq 2$$, $$g(x) = 2 - x^{2}$$.

$$
   \lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} (2 - x^{2}) = 2 - (1)^2 = 1
   $$

Since both the left-hand and right-hand limits as $$x$$ approaches 1 are equal to 1, the overall limit exists and is equal to 1, regardless of the value of $$g(1)$$ itself.

**Answer:**  
The limit is 1
The limit as $$ x $$ approaches 1 of $$ g(x) $$ is 1.

Let

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