Find the limit

[alexmay]

lim (square root of x ^2 +12x) -x
x-> infinity

I need help figuring out how to approach this problem

 

[MarkFL]

We are given:

[MATH]L=\lim_{x\to\infty}\left(\sqrt{x^2+12x}-x\right)[/MATH]
This is an indeterminate form. What does the limit look like if we rationalize the numerator?

 

[alexmay]

I rationalized the numerator and canceled out everything and am left with 12x

 

[Deleted member 4993]

I rationalized the numerator and canceled out everything and am left with 12x

What happened to the denominator? (you started with "1" in the denominator)

 

[alexmay]

I got 12/ sqrt of x ^2 + 12x after rationalizing the denominator and canceling

 

[MarkFL]

I would write:

[MATH](\sqrt{x^2+12x}-x)\cdot\frac{\sqrt{x^2+12x}+x}{\sqrt{x^2+12x}+x}=\frac{x^2+12x-x^2}{\sqrt{x^2+12x}+x}=\frac{12x}{\sqrt{x^2+12x}+x}[/MATH]
Now, let's divide numerator and denominator by \(x\):

[MATH]\frac{12}{\sqrt{1+\dfrac{12}{x}}+1}[/MATH]
Hence:

[MATH]L=\lim_{x\to\infty}\left(\frac{12}{\sqrt{1+\dfrac{12}{x}}+1}\right)=?[/MATH]

 

[alexmay]

why do you plug 1 in for x^2 and x?

 

[MarkFL]

why do you plug 1 in for x^2 and x?

[MATH]\frac{\dfrac{12x}{x}}{\dfrac{\sqrt{x^2+12x}+x}{x}}=\frac{12}{\sqrt{1+\dfrac{12}{x}}+1}[/MATH]

 

[alexmay]

Oh okay I see to now, do I flip the 12/x and add the ones with a common denominator?

 

[MarkFL]

No, now all you have to consider is that 12/x goes to zero as x goes to infinity...

 

[Steven G]

I got 12/ sqrt of x ^2 + 12x after rationalizing the denominator and canceling

No! The x's do not cancel out!