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The question:
Find the limit as x tends to 3 for $\displaystyle \frac{\frac{1}{x}-\frac{1}{3}}{x - 3}$
Normally I'd apply l'Hôpital's rule, but we haven't learnt that yet, so I assume they want us to do it algebraically. I'm not sure how to factor it so I can cancel the denominator (that's usually the case for these types of questions).
Any assistance is most welcome. Cheers!
Put the numerator over a common denominator and then cancel the obvious common factor in your expression.Quote:
Originally Posted by Glitch
By that, do you mean this?
$\displaystyle \frac{\frac{1}{x}}{x - 3} - \frac{\frac{1}{3}}{x-3}$
$\displaystyle \frac{1}{x(x-3)} - \frac{1}{3(x-3)}$
$\displaystyle \frac{1}{x^2-3x} - \frac{1}{3x-9}$
$\displaystyle \frac{6x -9 -x^2}{3x^3 - 18x^2 + 27x}$
Because that's also 0/0. (Worried)
Also, you can eliminate the fractions in the numerator by multiplying the entire expression byQuote:
Originally Posted by Glitch
$\displaystyle 1=\frac{x}{x}$
and
$\displaystyle 1=\frac{3}{3}$
to get
$\displaystyle \displaystyle\huge\frac{\frac{1}{x}-\frac{1}{3}}{x-3}$
$\displaystyle =\displaystyle\huge\frac{\frac{1}{x}-\frac{1}{3}}{x-3}\ \frac{x}{x}\ \frac{3}{3}$
$\displaystyle =\displaystyle\huge\frac{3-x}{3x(x-3)}=-\frac{x-3}{3x(x-3)}$
Nice one Archie! I didn't think to do that.
Mr Fantastic's idea is
$\displaystyle \displaystyle\huge\frac{\frac{3}{3}\frac{1}{x}-\frac{1}{3}\frac{x}{x}}{x-3}$
$\displaystyle =\displaystyle\huge\frac{\frac{3-x}{3x}}{x-3}$
Ahh, I see. Thank you.
Yet another way to do the same thing: multiply both numerator and denominator of $\displaystyle \frac{\frac{1}{x}-\frac{1}{3}}{x- 3}$ by 3x to get $\displaystyle \frac{3- x}{3x(x-3)}$. Now, for x not equal to 3, that is equal to $\displaystyle -\frac{1}{3x}$ and you can take the limit easily.
