## anonymous one year ago find the limit as x approaches 3 of (1)/(x-3)

1. anonymous

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

2. tkhunny

What does the denominator do as x approaches 3?

3. anonymous

@tkhunny it becomes 0...?

4. anonymous

please help! i have this and one more limit problem and i'm done and ive been studying all day and im exhausted and I just don't understand these.

5. jim_thompson5910

I suggest looking at a table of values specifically x values close to x = 3 like x = 3.1, x = 3.01, x = 3.001 and x = 2.99, x = 2.999, x = 2.9999, etc

6. jim_thompson5910

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7. anonymous

I already know that the limit doesnt exist I just need to know how to prove it

8. jim_thompson5910

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9. jim_thompson5910

as x gets closer to 3 from the left side, the value of f(x) heads off to negative infinity so the left hand limit is equal to -infinity $\Large \lim_{x \to 3^{-}}f(x) = -\infty$

10. jim_thompson5910

as x gets closer to 3 from the right side, the value of f(x) heads off to positive infinity the right hand limit is equal to +infinity $\Large \lim_{x \to 3^{+}}f(x) = +\infty$

11. jim_thompson5910

LHL = left hand limit RHL = right hand limit since $\Large LHL \ne RHL$ the limit at x = 3 itself does not exist the graph confirms this https://www.desmos.com/calculator/e09hafa0f1

12. anonymous

@jim_thompson5910 how would I prove this algebraically? is there a way to manipulate or sinplify 1/x-3?

13. jim_thompson5910

I don't know any way to prove it algebraically other than to use a table maybe

14. anonymous

@tkhunny do you know of any way to prove it algebraically?

15. tkhunny

x - 3 = 0 -- Solve.

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