limit at infinity

[renegade05]

OMG someone please help with this.

This is the problem:
\(\displaystyle \lim_{x \to -\infty}\frac{\sqrt{5x^2+7x}-3x}{7x-3}\)

This is what im doing:
\(\displaystyle =\lim_{x \to -\infty}\frac{(\sqrt{5x^2+7x}-3x)(\sqrt{5x^2+7x}+3x)}{(7x-3)(\sqrt{5x^2+7x}+3x)}\)

\(\displaystyle =\lim_{x \to -\infty}\frac{x(7-4x)}{(7x-3)(\sqrt{5x^2+7x}+3x)}\)

\(\displaystyle =\lim_{x \to -\infty}\frac{\frac{x(7-4x)}{x}}{(\frac{7x-3}{x})(\frac{\sqrt{5x^2+7x}}{x}+\frac{3x}{x})}\)

\(\displaystyle =\lim_{x \to -\infty}\frac{7-4x}{(7-\frac{3}{x})(-\sqrt{\frac{5x^2}{x^2}+\frac{7x}{x^2}}+3)}\)

\(\displaystyle =\lim_{x \to -\infty}\frac{7-4x}{(7-\frac{3}{x})(-\sqrt{5+\frac{7}{x}}+3)}\)

So I am left wtih:
\(\displaystyle =\lim_{x \to -\infty}\frac{\infty}{(7)(-\sqrt{5}+3)}=\infty\)

I know this is not the answer because i plugged the expression into my graphing calc and put in values really large and negative and it appears the limit is approaching -.748009 which i do realize is \(\displaystyle \frac{-4}{(7)(-\sqrt{5}+3)}\)
BUT WHY??? i have checked, re-checked and then checked once more! Where am i screwing up? Something fundamental? some arthimitic ? algebra?
I have no idea what i am doing wrong. please help me.

 

[galactus]

\(\displaystyle \lim_{x\to \infty}\frac{\sqrt{5x^{2}+7x}-3x}{7x-3}\)

It would be helpful to manipulate the function so that powers of x becomes powers of 1/x. We can do this by dividing

the numerator and denominator by |x| and using the fact that \(\displaystyle \sqrt{x^{2}}=|x|\)

As \(\displaystyle x\to \infty\), the values of x are eventually positive, so we replace |x| by x where we want.

Doing so leads to:

\(\displaystyle \lim_{x\to \infty}\frac{\sqrt{\frac{5x^{2}}{x^{2}}+\frac{7}{x}}-\frac{3x}{x}}{\frac{\frac{7x}{x}-\frac{3}{x}}}\)

We are left with:

\(\displaystyle \frac{\sqrt{5}-3}{7}\)

 

[renegade05]

so since x is approaching \(\displaystyle -\infty\)

the answer would be:
\(\displaystyle \frac{-\sqrt{5}-3}{7}\)

????

 

[mmm4444bot]

Yes

That's about -0.7480

 

[daon]

Your error was when dividing by x. You divided the numerator once by x, but the denominator TWICE by x.

 

[galactus]

so since x is approaching \(\displaystyle -\infty\)

the answer would be:
\(\displaystyle \frac{-\sqrt{5}-3}{7}\)

????

Yes. Sorry about that. I did not notice the negative on the infinity in your original problem.

 

[renegade05]

Your error was when dividing by x. You divided the numerator once by x, but the denominator TWICE by x.

I dont think I did....

I multiplied by the conjugate of the numerator which was very unnecessary. But i only divided once...