limit of (9x^2+3)^1/2 - (3x) as x approches neg infinity
- Thread starter member566
- Start date
Re: limit as x approches neg infinity
Hello, member566!
You started off correctly . . .
We have: \(\displaystyle \L\:\frac{x}{\sqrt{9x^2\,+\,x}\,+\,3x}\)
Divide top and bottom by \(\displaystyle x\)
\(\displaystyle \L\frac{\frac{x}{x}}{\frac{\sqrt{9x^2\,+\,x}}{x}\,+\,\frac{x}{x}} \;=\;\frac{1}{\frac{\sqrt{9x^2\,+\,x}}{\sqrt{x^2}}\,+\,1} \;=\;\frac{1}{\sqrt{\frac{9x^2\,+\,x}{x^2}} \,+\,1} \;=\;\frac{1}{\sqrt{9\,+\,\frac{1}{x}}\,+\,1}\)
Then: \(\displaystyle \L\:\lim_{x\to-\infty}\,\frac{1}{\sqrt{9\,+\,\frac{1}{x}}\,+\,1} \;=\;\frac{1}{\sqrt{9\,+\,0}\,+\,1} \;=\;\frac{1}{3\,+\,1}\;=\;\fbox{\frac{1}{4}}\)
Hello, member566!
You started off correctly . . .
We have: \(\displaystyle \L\:\frac{x}{\sqrt{9x^2\,+\,x}\,+\,3x}\)
Divide top and bottom by \(\displaystyle x\)
\(\displaystyle \L\frac{\frac{x}{x}}{\frac{\sqrt{9x^2\,+\,x}}{x}\,+\,\frac{x}{x}} \;=\;\frac{1}{\frac{\sqrt{9x^2\,+\,x}}{\sqrt{x^2}}\,+\,1} \;=\;\frac{1}{\sqrt{\frac{9x^2\,+\,x}{x^2}} \,+\,1} \;=\;\frac{1}{\sqrt{9\,+\,\frac{1}{x}}\,+\,1}\)
Then: \(\displaystyle \L\:\lim_{x\to-\infty}\,\frac{1}{\sqrt{9\,+\,\frac{1}{x}}\,+\,1} \;=\;\frac{1}{\sqrt{9\,+\,0}\,+\,1} \;=\;\frac{1}{3\,+\,1}\;=\;\fbox{\frac{1}{4}}\)
Not quite there. Pretend the \(\displaystyle \lim_{x \to -\infty}\) is applied to each line:
\(\displaystyle \L\frac{x}{\sqrt{9x^{2} + x} + 3x} \cdot \frac{\;\frac{1}{x}\;}{\;\frac{1}{x}\;}\)
\(\displaystyle \L= \frac{1}{\frac{\sqrt{9x^{2} + x}}{x} + \frac{3x}{x}}\)
\(\displaystyle \L= \frac{1}{\frac{\sqrt{9x^{2} + x}}{-\sqrt{x^{2}}} + 3}\) (since x < 0 then \(\displaystyle \sqrt{x^{2}} = |x| = -x\))
\(\displaystyle \L= \frac{1}{-\sqrt{\frac{9x^{2} + x}{x^{2}}} + 3}\)
\(\displaystyle \L= \frac{1}{-\sqrt{9 + \frac{1}{x}} + 3}\)
\(\displaystyle \L= \frac{1}{-3 + 3} = +\infty\)
\(\displaystyle \L\frac{x}{\sqrt{9x^{2} + x} + 3x} \cdot \frac{\;\frac{1}{x}\;}{\;\frac{1}{x}\;}\)
\(\displaystyle \L= \frac{1}{\frac{\sqrt{9x^{2} + x}}{x} + \frac{3x}{x}}\)
\(\displaystyle \L= \frac{1}{\frac{\sqrt{9x^{2} + x}}{-\sqrt{x^{2}}} + 3}\) (since x < 0 then \(\displaystyle \sqrt{x^{2}} = |x| = -x\))
\(\displaystyle \L= \frac{1}{-\sqrt{\frac{9x^{2} + x}{x^{2}}} + 3}\)
\(\displaystyle \L= \frac{1}{-\sqrt{9 + \frac{1}{x}} + 3}\)
\(\displaystyle \L= \frac{1}{-3 + 3} = +\infty\)
I was just looking at the problem and all the answers were positive, yet the limit is in the neg inf. Should not the answer allso be neg, some one allso posted that neg infinity should be pluged in for every X, this works for a limit problem with an inteber, however, pluging in inf there is no real answer.
need more info.
Thank you.
need more info.
Thank you.
Well, look at your equation again:
\(\displaystyle \sqrt{9x^{2} + x} - 3x\)
We know x is always negative but if you look at the equation, the result will always be positive. In the square root portion, 9x<sup>2</sup> > x which will give you a positive value. \(\displaystyle -3x\) will be positive because it's the product of two negative numbers.
\(\displaystyle \sqrt{9x^{2} + x} - 3x\)
We know x is always negative but if you look at the equation, the result will always be positive. In the square root portion, 9x<sup>2</sup> > x which will give you a positive value. \(\displaystyle -3x\) will be positive because it's the product of two negative numbers.