What is the limit as x approaches 2 of the function (1/x - 1/3)/(x-3)? The fractions part of the problem made me confused and get the wrong answer.

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What is the limit as x approaches 2 of the function (1/x - 1/3)/(x-3)? The fractions part of the problem made me confused and get the wrong answer.

Calculus1
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\[\lim_{x \rightarrow 3} (\frac{ 1 }{ x } - \frac{ 1 }{ 3 })/(x-3) \]
what would you get for.. say \(\bf \cfrac{1}{x}-\cfrac{1}{3}?\)
You'd get 0 when you subsitute 3 for x.

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well.. I meant the numerator only :)
Oh ok. :D
The only thing I don't get is that I got 0 both in the numerator and the denominator. Not sure what to do next.
right... one sec bear in mind that \(\bf a-b \iff -(b-a)\) one sec
Ok
\(\bf \lim\limits_{x\to 3}\ \cfrac{\frac{1}{x}-\frac{1}{3}}{x-3} \\ \quad \\ \cfrac{\frac{3-x}{3x}}{x-3}\implies \cfrac{\frac{3-x}{3x}}{\frac{x-3}{1}}\implies \cfrac{3-x}{3x}\cdot \cfrac{1}{x-3} \\ \quad \\ \cfrac{3-x}{3x(x-3)}\implies \cfrac{{\color{brown}{ -\cancel{(x-3)}}}}{3x\cancel{(x-3)}}\)
Oh, now I see what I was supposed to do! Thank you very much!
yw

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