Question

construct a table and find the indicated limit.
if h(x)= \frac{\sqrt{x} + 2}{x - 2}, then find \lim\limits_{x\to 2} h(x).
complete the table below.
x \quad 1.9 \quad 1.99 \quad 1.999 \quad 2.001 \quad 2.01 \quad 2.1
h(x) \quad -33.7840 \quad -341.0674 \quad -3413.8600 \quad 3414.5671 \quad 341.7745 \quad 34.4914
(type integers or decimals rounded to four decimal places as needed.)
what is the limit? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
\bigcirc a. \quad \lim\limits_{x\to 2} \frac{\sqrt{x} + 2}{x - 2} = \square
\bigcirc b. the limit does not exist and is neither \infty nor -\infty.

Explanation:

Step1: Analyze left - hand limit

As \(x\) approaches \(2\) from the left (\(x = 1.9,1.99,1.999\)), the values of \(h(x)\) are \(- 33.7840,-341.0674,-3413.8600\). These values are getting more and more negative (tending to \(-\infty\)).

Step2: Analyze right - hand limit

As \(x\) approaches \(2\) from the right (\(x = 2.001,2.01,2.1\)), the values of \(h(x)\) are \(3414.5671,341.7745,34.4914\). These values are getting more and more positive (tending to \(+\infty\)).

Step3: Compare left and right limits

For the limit \(\lim_{x
ightarrow a}f(x)\) to exist, the left - hand limit (\(\lim_{x
ightarrow a^{-}}f(x)\)) and the right - hand limit (\(\lim_{x
ightarrow a^{+}}f(x)\)) must be equal. Here, \(\lim_{x
ightarrow 2^{-}}h(x)=-\infty\) and \(\lim_{x
ightarrow 2^{+}}h(x)=+\infty\), so the two one - sided limits are not equal. Also, the function values do not approach a finite number.

Answer:

B. The limit does not exist and is neither \(\infty\) nor \(-\infty\).