I came across 3 problems where the only difference between them is the sign of the value to which $x$ is approaching to. Please see below:

$$ 1) \lim_{x\to3^-}\frac{x}{x-3} $$ $$ 2) \lim_{x\to3^+}\frac{x}{x-3} $$ $$ 3) \lim_{x\to3}\frac{x}{x-3} $$

At least in example 3 we can't use limits arithmetic because if $x=3$ the denominator equals $0$. In that case we should somehow simplify the expression which I don't really see how. In the cases 1 and 2 I feel that the sign can help with simpler solution but not sure how exactly.


First solve $1$ and $2$, and from that, you can deduce the answer to $3$ (i.e., the limit in $3$ exists if and only if the limits in $1$ and $2$ exist and are equal).

For $1$ and $2$, the sign makes it easier, yes. Take a look at $1$, where, since $x\to 3^-$, you know that $x<3$ and therefore that $x-3<0$.

So, you know that $\frac{x}{x-3}$ will be negative, since $x-3$ is negative (and $x>0$ for $x$ close enough to $3$).

Now, to actually calculate the limit, a hint:

  • If $x$ is very close to $3$, what happens to $x-3$?
  • What, then, happens to $\frac{1}{|x-3|}$?
  • What about $\frac1{x-3}$ (remember, $x\to 3^-$)?
  • $\begingroup$ sorry the hint didn't help. I just don't see how to simplify the expression, otherwise it's almost $0$ in the denominator. By the way, the problems were unrelated. $\endgroup$ – Yos Nov 15 '16 at 8:31
  • 1
    $\begingroup$ @Yos You don't have to simplify the expression. You have to find the limit. Also, please tell me your answers to the questions in the hint. $\endgroup$ – 5xum Nov 15 '16 at 8:36
  • $\begingroup$ If $x$ is very close to $3$, then $x-3$ is very close to $0$. $\endgroup$ – Yos Nov 15 '16 at 8:39
  • $\begingroup$ @Yos And then what happens to $\frac{1}{|x-3|}$? $\endgroup$ – 5xum Nov 15 '16 at 8:40
  • $\begingroup$ it's close to $0$ as well but we can't have $0$ in the denominator $\endgroup$ – Yos Nov 15 '16 at 8:40

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