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$g(x) = 2^{x} - 1$ for $-8 \leq x < 1$ ; $\sqrt{x}$ for $x \geq 1$ $\newline$ Find $\lim_{x \to 4} g(x)$ .

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Find derivatives using the quotient rule II

Full solution

Q.

g(x) = 2^{x} - 1

for

-8 \leq x < 1

;

\sqrt{x}

for

x \geq 1

\newline

Find

\lim_{x \to 4} g(x)

.

Identify Function Piece: Since $x$ is approaching $4$ , we look at the piece of the function that applies when $x$ is greater than or equal to $1$ , which is $\sqrt{x}$ .
Substitute $x=4$ : Now we just plug in $4$ into $\sqrt{x}$ to find the limit.
Calculate Limit: So, $\lim_{x \to 4} g(x) = \sqrt{4}.$
Final Result: Calculating $\sqrt{4}$ gives us $2$ .

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