- #1
fedaykin
- 138
- 3
1. My problem is such:
Find the limit of [tex]\lim_{x \rightarrow \infty} \sqrt{9x^2+x} -3x [/tex]
2. No relevant equations
3. I multiplied [tex] \frac{\sqrt{9x^2+x} -3x}{1} * \frac{\sqrt{9x^2+x} +3x}{\sqrt{9x^2+x} +3x} = \frac{x}{\sqrt{9x^2+x} +3x} [/tex]
I am now quite confused as to where to go from here. My teacher will accept tabled values, but I wish to prove my answer. I would greatly appreciate any help.
I did attempt to divide by the highest power in the denominator, but all that got me was a mess:
[tex] \frac{1}{\frac{\sqrt{9x^2+x}}{x} +3} [/tex]
Ok, I'm not certain this is valid, but...:
I'll factor out an x under the radical, then attempt to simplify
[tex] \frac{1}{\frac{\sqrt{9x^2*(1+\frac}{x}{9x^2}{x} +3} [/tex]
Hmm... I'm having trouble with tex and that.
Find the limit of [tex]\lim_{x \rightarrow \infty} \sqrt{9x^2+x} -3x [/tex]
2. No relevant equations
3. I multiplied [tex] \frac{\sqrt{9x^2+x} -3x}{1} * \frac{\sqrt{9x^2+x} +3x}{\sqrt{9x^2+x} +3x} = \frac{x}{\sqrt{9x^2+x} +3x} [/tex]
I am now quite confused as to where to go from here. My teacher will accept tabled values, but I wish to prove my answer. I would greatly appreciate any help.
I did attempt to divide by the highest power in the denominator, but all that got me was a mess:
[tex] \frac{1}{\frac{\sqrt{9x^2+x}}{x} +3} [/tex]
Ok, I'm not certain this is valid, but...:
I'll factor out an x under the radical, then attempt to simplify
[tex] \frac{1}{\frac{\sqrt{9x^2*(1+\frac}{x}{9x^2}{x} +3} [/tex]
Hmm... I'm having trouble with tex and that.
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