hpfan101
  • hpfan101
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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SOLVED
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chestercat
  • chestercat
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hpfan101
  • hpfan101
\[\lim_{x \rightarrow 3} (\frac{ 1 }{ x } - \frac{ 1 }{ 3 })/(x-3) \]
jdoe0001
  • jdoe0001
what would you get for.. say \(\bf \cfrac{1}{x}-\cfrac{1}{3}?\)
hpfan101
  • hpfan101
You'd get 0 when you subsitute 3 for x.

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jdoe0001
  • jdoe0001
well.. I meant the numerator only :)
hpfan101
  • hpfan101
Oh ok. :D
hpfan101
  • hpfan101
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.
jdoe0001
  • jdoe0001
right... one sec bear in mind that \(\bf a-b \iff -(b-a)\) one sec
hpfan101
  • hpfan101
Ok
jdoe0001
  • jdoe0001
\(\bf \lim\limits_{x\to 3}\ \cfrac{\frac{1}{x}-\frac{1}{3}}{x-3} \\ \quad \\ \cfrac{\frac{3-x}{3x}}{x-3}\implies \cfrac{\frac{3-x}{3x}}{\frac{x-3}{1}}\implies \cfrac{3-x}{3x}\cdot \cfrac{1}{x-3} \\ \quad \\ \cfrac{3-x}{3x(x-3)}\implies \cfrac{{\color{brown}{ -\cancel{(x-3)}}}}{3x\cancel{(x-3)}}\)
hpfan101
  • hpfan101
Oh, now I see what I was supposed to do! Thank you very much!
jdoe0001
  • jdoe0001
yw

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