$$
g(x)=\left\{\begin{array}{ll}
2^{x}-1 & \text { for }-8 \leq x

Q: \[ g(x)=\left\{\begin{array}{ll} 2^{x}-1 & \text { for }-8 \leq x<1 \\ \sqrt{x} & \text { for } x \geq 1 \end{array}\right. \] Find $\lim _{x \rightarrow 4} g(x)$. Choose 1 answer: (A) 1 (B) 2 (c) 15 D The limit doesn't exist.

∻Solution by Steps∻ ‖ step 1 ⋮ Identify the piece of the function that is relevant for the limit as $ x $ approaches 4. ‖ ‖ step 2 ⋮ Since $ x \geq 1 $ in the second piece of the function, and 4 is greater than 1, we use $ \sqrt{x} $ to find the limit. ‖ ‖ step 3 ⋮ Calculate the limit using the square root function: $ \lim_{x \to 4} \sqrt{x} $. ‖ ‖ step 4 ⋮ Substitute $ x $ with 4 in the square root function to find the limit: $ \sqrt{4} = 2 $. ‖ ∻Answer∻ ⚹ (B) 2 ⚹ ∻Key Concept∻ ⚹Piecewise Function Limits⚹ ∻Explanation∻ ⚹The limit of a piecewise function as $ x $ approaches a value is determined by the piece of the function that is relevant for that value of $ x $. In this case, since 4 is greater than 1, we use the second piece, $ \sqrt{x} $, to find the limit. The limit as $ x $ approaches 4 of $ \sqrt{x} $ is 2.⚹◊What is the formula to find the limit of a piecewise function like $g(x)$ as $x$ approaches a specific value?⍭ Generate me a similar question◊

Question

$$
g(x)=\left\{\begin{array}{ll}
2^{x}-1 & \text { for }-8 \leq x<1 \
\sqrt{x} & \text { for } x \geq 1
\end{array}\right.
$$

Find $\lim _{x \rightarrow 4} g(x)$.
Choose 1 answer:
(A) 1
(B) 2
(c) 15

D The limit doesn't exist.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Identify the piece of the function that is relevant for the limit as $ x $ approaches 4. ‖ ‖ step 2 ⋮ Since $ x \geq 1 $ in the second piece of the function, and 4 is greater than 1, we use $ \sqrt{x} $ to find the limit. ‖ ‖ step 3 ⋮ Calculate the limit using the square root function: $ \lim_{x \to 4} \sqrt{x} $. ‖ ‖ step 4 ⋮ Substitute $ x $ with 4 in the square root function to find the limit: $ \sqrt{4} = 2 $. ‖ ∻Answer∻ ⚹ (B) 2 ⚹ ∻Key Concept∻ ⚹Piecewise Function Limits⚹ ∻Explanation∻ ⚹The limit of a piecewise function as $ x $ approaches a value is determined by the piece of the function that is relevant for that value of $ x $. In this case, since 4 is greater than 1, we use the second piece, $ \sqrt{x} $, to find the limit. The limit as $ x $ approaches 4 of $ \sqrt{x} $ is 2.⚹◊What is the formula to find the limit of a piecewise function like $g(x)$ as $x$ approaches a specific value?⍭ Generate me a similar question◊

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