Question

for the function g(x) shown below, compute the following limits or state that they do not exist.

g(x) = \begin{cases} 0 & \text{if } x leq -5 \\sqrt{25 - x^2} & \text{if } -5 < x < 5 \\ 2x & \text{if } x geq 5 end{cases}

a. (limlimits_{x \to -5^-} g(x)) b. (limlimits_{x \to -5^+} g(x)) c. (limlimits_{x \to -5} g(x)) d. (limlimits_{x \to 5^-} g(x)) e. (limlimits_{x \to 5^+} g(x)) f. (limlimits_{x \to 5} g(x))

...

a. (limlimits_{x \to 5^+} g(x) = 10)

b. the limit does not exist.

f. find (limlimits_{x \to 5} g(x)). select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. (limlimits_{x \to 5} g(x) = square)

b. the limit does not exist.

Explanation:

Step1: Find left - hand limit as \(x

ightarrow5^{-}\)
For \(x
ightarrow5^{-}\), we use the piece of the function \(g(x)=\sqrt{25 - x^{2}}\) (since \(- 5\lt x\lt5\) for this piece and as \(x\) approaches \(5\) from the left, \(x\) is in the interval \(-5\lt x\lt5\)).
We calculate \(\lim_{x
ightarrow5^{-}}g(x)=\lim_{x
ightarrow5^{-}}\sqrt{25 - x^{2}}\). Substitute \(x = 5\) into \(\sqrt{25 - x^{2}}\), we get \(\sqrt{25-5^{2}}=\sqrt{25 - 25}=0\).

Step2: Find right - hand limit as \(x

ightarrow5^{+}\)
For \(x
ightarrow5^{+}\), we use the piece of the function \(g(x) = 2x\) (since \(x\geq5\) for this piece and as \(x\) approaches \(5\) from the right, \(x\) is in the interval \(x\geq5\)).
We calculate \(\lim_{x
ightarrow5^{+}}g(x)=\lim_{x
ightarrow5^{+}}2x\). Substitute \(x = 5\) into \(2x\), we get \(2\times5 = 10\).

Step3: Determine the existence of \(\lim_{x

ightarrow5}g(x)\)
For the limit \(\lim_{x
ightarrow a}f(x)\) to exist, we need \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)\).
We found that \(\lim_{x
ightarrow5^{-}}g(x) = 0\) and \(\lim_{x
ightarrow5^{+}}g(x)=10\). Since \(0
eq10\), the two - sided limit \(\lim_{x
ightarrow5}g(x)\) does not exist.

Answer:

B. The limit does not exist.