Check Limit at Infinity of f(x)

[Zhalfirin88]

Homework Statement

Can you guys just check to see if I'm right?

f(x) = \frac{2x-\sqrt{4x^2-5x+300}}{1}

The Attempt at a Solution

\frac{2x - \sqrt{4x^2-5x+300}}{1} * \frac{2x+ \sqrt{4x^2-5x+300}}{2x+ \sqrt{4x^2-5x+300}}}

\frac{4x^2 - 4x^2 + 5x - 300}{2x+ \sqrt{4x^2-5x+300}}

\frac{\frac{5x-300}{x}}{\frac{2x+ \sqrt{4x^2-5x+300}}{x}}

\frac{5- \frac{300}{x}}{2 + \sqrt{ \frac{4x^2 - 5x + 300}{x^2}}}

\frac{5 + 0}{2 + \sqrt{ \frac{4x^2}{x^2} - \frac{5x}{x^2} + \frac{300}{x^2}}}

\frac{5}{2 + \sqrt{4 - 0 +0}}

\frac{5}{4}

 

[emyt]

Homework Statement

Can you guys just check to see if I'm right?

f(x) = \frac{2x-\sqrt{4x^2-5x+300}}{1}

I don't know how to do Latex for limits at infinity, but the question is find the limit of f(x) as x approaches infinity.

The Attempt at a Solution

\frac{2x - \sqrt{4x^2-5x+300}}{1} * \frac{2x+ \sqrt{4x^2-5x+300}}{2x+ \sqrt{4x^2-5x+300}}}

\frac{4x^2 - 4x^2 + 5x - 300}{2x+ \sqrt{4x^2-5x+300}}

\frac{\frac{5x-300}{x}}{\frac{2x+ \sqrt{4x^2-5x+300}}{x}}

\frac{5- \frac{300}{x}}{2 + \sqrt{ \frac{4x^2 - 5x + 300}{x^2}}}

\frac{5 + 0}{2 + \sqrt{ \frac{4x^2}{x^2} - \frac{5x}{x^2} + \frac{300}{x^2}}}

\frac{5}{2 + \sqrt{4 - 0 +0}}

\frac{5}{4}

yes, it's 5/4.. generally, the technique is to divide out by the term with the highest degree

 

[tiny-tim]

I don't know how to do Latex for limits at infinity, but the question is find the limit of f(x) as x approaches infinity.

Hi Zhalfirin88!

It's \lim_{x\rightarrow \infty} … \lim_{x\rightarrow \infty}

 

[Zhalfirin88]

Oh, thanks tiny-tim :)