limit x = infinity ((sqrt9x^2+x)-3x)

[Sophie]

Could someone please explain why my method differs from the worked method.

I am trying to find the limit x = infinity ((sqrt9x^2+x)-3x)

I divided everything by x

= sqrt(9+1/x) – 3

=Sqrt(lim9 +lim1/x) – lim3

=Sqrt (9+0) – 3

=3-3 = 0

The worked example multiplies (sqrt9x^2+x)+3x to the numerator and denominator. I understand this, but I do not understand why mine does not work as well. Is it something to do with the brackets? If someone could explain that would be wonderful.

Thanks Sophie

 

[stapel]

I divided everything by x

Since, for instance, 3x does not equal 3, "dividing by x" cannot be a legitimate step.

(It's different with fractions, of course, since multiplying by (1/x)/(1/x) is just multiplying by a form of "1", a legitimate mathematical operation.)

Eliz.

 

[soroban]

Hello, Sophie!

I am trying to find: \(\displaystyle \L\lim_{x\to\infty}\left(\sqr{9x^2\,+\,x}\,-\,3x\right)\)

You can't simply "divide by \(\displaystyle x\)" like that . . . You've changed the problem.


Multiply top and bottom by the conjugate:

. . \(\displaystyle \L\frac{\sqrt{9x^2+x}\,-\,3x}{1}\,\cdot\,\frac{\sqrt{9x^2+x}\,+\,3x}{\sqrt{9x^2+x}\,+\,3x} \;=\;\frac{(9x^2+x)\,-\,9x^2}{\sqrt{9x^2+x}\,+\,3x} \;=\;\frac{x}{\sqrt{9x^2+x}\,+\,3x}\)


Divide top and bottom by \(\displaystyle x:\)

. . \(\displaystyle \L\frac{\frac{x}{x}} {\frac{\sqrt{x^2+x}}{x} + \frac{3x}{x}} \;=\;\frac{1}{\sqrt{\frac{9x^2+x}{x^2}} \,+\,3} \;=\;\frac{1}{\sqrt{9\,+\,\frac{1}{x}}\,+\,3}\)


Take the limit: \(\displaystyle \L\:\lim_{x\to\infty}\left[\frac{1}{\sqrt{9\,+\,\frac{1}{x}}\,+\,3}\right] \;=\;\frac{1}{\sqrt{9\,+\,0}\,+\,3} \;=\;\frac{1}{3\,+\,3}\;=\;\fbox{\frac{1}{6}}\)