Limits with radicals

[temporaryinsanit]

I'm having a problem finding the answer to these questions, I know I have to rationalize the expression and go on from there but I still get stuck. Can someone tell the steps to take to solve each problem and how to get the answer. Thank You

A. lim 2 - ?4 – x / x
(x?0)

B. lim 3x / ?x² - 4
(x??)

 

[soroban]

Hello, temporaryinsanit!

These are two different types . . .

\(\displaystyle (A)\;\;\lim_{x\to0}\frac{2 - \sqrt{4-x} }{ x }\)

Multiply top and bottom by the conjugate:

. . \(\displaystyle \frac{2-\sqrt{4-x}}{x}\cdot\frac{2+\sqrt{4-x}}{2+\sqrt{4 - x}} \;=\;\frac{4 - (4-x)}{x(2+\sqrt{4-x})}\)

. . . . \(\displaystyle =\;\frac{x}{x(2 + \sqrt{4-x})} \;=\; \frac{1}{2+\sqrt{4-x}}\)


Now take the limit:

. . \(\displaystyle \lim_{x\to0}\frac{1}{2+\sqrt{4-x}} \;=\;\frac{1}{2 + \sqrt{4}} \;=\;\frac{1}{2+2} \;=\;\frac{1}{4}\)

\(\displaystyle (B)\;\;\lim_{x\to\infty}\frac{3x}{\sqrt{x^2 - 4}}\)

Divide top and bottom by \(\displaystyle x:\)

. . \(\displaystyle \frac{\dfrac{3x}{x}}{\dfrac{\sqrt{x^2-4}}{x}} \;=\;\frac{3}{\dfrac{\sqrt{x^2-4}}{\sqrt{x^2}}} \;=\;\frac{3}{\sqrt{\dfrac{x^2-4}{x^2}}} \;=\; \frac{3}{\sqrt{1-\frac{4}{x^2}}}\)


Now take the limit:

. . \(\displaystyle \lim_{x\to\infty}\frac{3}{\sqrt{1-\frac{4}{x^2}}} \;=\; \frac{3}{\sqrt{1-0}} \;=\;3\)