Math Help - Finding a limit

1. 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!

2. 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.

3. 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.

4. 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

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

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

$=\displaystyle\huge\frac{3-x}{3x(x-3)}=-\frac{x-3}{3x(x-3)}$

5. Nice one Archie! I didn't think to do that.

6. Mr Fantastic's idea is

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

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

7. Ahh, I see. Thank you.

8. 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.