Finding a limit

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August 1st 2010, 01:00 AM
Glitch

Finding a limit

The question:
Find the limit as x tends to 3 for $\frac{\frac{1}{x}-\frac{1}{3}}{x - 3}$

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!
August 1st 2010, 01:03 AM
mr fantastic

Quote:

Originally Posted by Glitch

The question:
Find the limit as x tends to 3 for $\frac{\frac{1}{x}-\frac{1}{3}}{x - 3}$

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.
August 1st 2010, 03:00 AM
Glitch

By that, do you mean this?

$\frac{\frac{1}{x}}{x - 3} - \frac{\frac{1}{3}}{x-3}$

$\frac{1}{x(x-3)} - \frac{1}{3(x-3)}$

$\frac{1}{x^2-3x} - \frac{1}{3x-9}$

$\frac{6x -9 -x^2}{3x^3 - 18x^2 + 27x}$

Because that's also 0/0. (Worried)
August 1st 2010, 03:00 AM
Archie Meade

Quote:

Originally Posted by Glitch

The question:
Find the limit as x tends to 3 for $\frac{\frac{1}{x}-\frac{1}{3}}{x - 3}$

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!

Also, you can eliminate the fractions in the numerator by multiplying the entire expression by

$1=\frac{x}{x}$

and

$1=\frac{3}{3}$

to get

$\displaystyle\huge\frac{\frac{1}{x}-\frac{1}{3}}{x-3}$

$=\displaystyle\huge\frac{\frac{1}{x}-\frac{1}{3}}{x-3}\ \frac{x}{x}\ \frac{3}{3}$

$=\displaystyle\huge\frac{3-x}{3x(x-3)}=-\frac{x-3}{3x(x-3)}$
August 1st 2010, 03:06 AM
Glitch

Nice one Archie! I didn't think to do that.
August 1st 2010, 03:15 AM
Archie Meade

Mr Fantastic's idea is

$\displaystyle\huge\frac{\frac{3}{3}\frac{1}{x}-\frac{1}{3}\frac{x}{x}}{x-3}$

$=\displaystyle\huge\frac{\frac{3-x}{3x}}{x-3}$
August 1st 2010, 03:19 AM
Glitch

Ahh, I see. Thank you.
August 1st 2010, 04:19 AM
HallsofIvy

Yet another way to do the same thing: multiply both numerator and denominator of $\frac{\frac{1}{x}-\frac{1}{3}}{x- 3}$ by 3x to get $\frac{3- x}{3x(x-3)}$ . Now, for x not equal to 3, that is equal to $-\frac{1}{3x}$ and you can take the limit easily.