Question

I attached a picture of the practice problem I'm trying to solve

Answer 1

Explanation

From the statement, we know that:

$\begin{gathered} f(2)=0,\text{ }g(2)=-1, \\ \lim_{x\to2}\text{ }\lbrack f(x)+g(x)\rbrack=7, \\ \lim_{x\to2}\text{ }\lbrack2f(x)-g(x)\rbrack=-1. \end{gathered}$

(a) Summing the given limits, applying the distributive property for the limit, we get:

$\begin{gathered} \lim_{x\to2}\text{ }\lbrack f(x)+g(x)\rbrack+\lim_{x\to2}\text{ }\lbrack2f(x)-g(x)\rbrack=7-1, \\ 3\lim_{x\to2}f(x)=6,\text{ } \\ \lim_{x\to2}f(x)=\frac{6}{3}=2. \end{gathered}$

Using this result in the first given limit, we get:

$\begin{gathered} \lim_{x\to2}\lbrack f(x)+g(x)]=7, \\ \lim_{x\to2}f(x)+\lim_{x\to2}g(x)=7, \\ 2+\lim_{x\to2}g(x)=7, \\ \lim_{x\to2}g(x)=7-2=5. \end{gathered}$

(b) Putting the limit inside the square root and in the denominator, and then using the results from above, we get:

$\begin{gathered} \lim_{x\to2}\lbrack\frac{\sqrt{g(x)+4}}{f(x)}] \\ =\frac{\lim_{x\to2}\lbrack\sqrt{g(x)+4}]}{\lim_{x\to2}f(x)} \\ =\frac{\sqrt{\lim_{x\to2}g(x)+4}}{\lim_{x\to2}f(x)} \\ =\frac{\sqrt{5+4}}{2} \\ =\frac{3}{2}. \end{gathered}$

(c) Putting the limit inside the exponent of the e power, and using the results from above, we get:

$\lim_{x\to2}\lbrack e^{f(x)-11}\rbrack=e^{\lim_{x\to2}f(x)-11}=e^{2-11}=e^{-9}.$

(d) Using the results from above, we get:

$\begin{gathered} \lim_{x\to2}\lbrack(f(x))^3-2g(x)+22] \\ =(\lim_{x\to2}f(x))^3-2\lim_{x\to2}g(x)+\lim_{x\to2}22 \\ =2^3-2\cdot5+22 \\ =20. \end{gathered}$Answer

a. 5

b. 3/2

c. e^(-9)

d. 20