limits with infinity

[Dorian Gray]

Greetings Mathematicians,

I am having troubles with this problem. I do not know where to start with this problem. Should I use a conjugate? Divide by the highest exponent from the denominator?

 

[pka]

Greetings Mathematicians,

I am having troubles with this problem. I do not know where to start with this problem. Should I use a conjugate? Divide by the highest exponent from the denominator?View attachment 1634

If \(\displaystyle x>0\) then
\(\displaystyle \dfrac{x+2}{\sqrt{9x^2+1}}=\dfrac{1+\frac{2}{x}}{\sqrt{9+\frac{1}{x^2}}}\)

 

[Dorian Gray]

thank you

Thank you PKA for your prompt response. I see that you divided by the highest exponent in the denominator. It appears then that the limit is 1/3....

 

[soroban]

Hello, Dorian Gray!

\(\displaystyle \displaystyle \lim_{x\to\infty}\,\frac{x+2}{\sqrt{9x^2+1}}\)

We have: .\(\displaystyle \dfrac{x+2}{\sqrt{9x^2+1}}\)


Divide numerator and denominator by \(\displaystyle x\!:\)

. . \(\displaystyle \dfrac{\dfrac{x+2}{x}}{\dfrac{\sqrt{9x^2+1}}{x}} \;=\;\dfrac{\dfrac{x}{x} + \dfrac{2}{x}} {\dfrac{\sqrt{9x^2+1}}{\sqrt{x^2}}} \;=\;\dfrac{1 + \dfrac{2}{x}}{\sqrt{\dfrac{9x^2+1}{x^2}}} \;=\;\dfrac{1 + \dfrac{2}{x}}{\sqrt{\dfrac{9x^2}{x^2} + \dfrac{1}{x^2}}} \;=\;\dfrac{1 + \dfrac{2}{x}}{\sqrt{9 + \dfrac{1}{x^2}}} \)


\(\displaystyle \displaystyle\text{Therefore: }\:\lim_{x\to\infty}\frac{1+\frac{2}{x}}{\sqrt{9 + \frac{1}{x^2}}} \;=\;\frac{1 + 0}{\sqrt{9 + 0}} \;=\;\frac{1}{3} \)