infinite limit

• Aug 29th 2009, 03:19 PM
DBA
infinite limit
Hello,
I do not understand this...

Determine the infinite limit:

lim x-->-3+ f(x); f(x) = x+2/x+3

I looked at the numerator and denominator...

(x+2) approaches -1 as x gets closer to -3 from the right and
(x+3) approchaes 0 as x gets closer to -3 from the right

How do I know that the function f(x) has an limit of negative infinity?
How do I know that the limit is infinite at all? (Why not just "does not exist")

Thanks
• Aug 29th 2009, 05:20 PM
mr fantastic
Quote:

Originally Posted by DBA
Hello,
I do not understand this...

Determine the infinite limit:

lim x-->-3+ f(x); f(x) = x+2/x+3

I looked at the numerator and denominator...

(x+2) approaches -1 as x gets closer to -3 from the right and
(x+3) approchaes 0 as x gets closer to -3 from the right

How do I know that the function f(x) has an limit of negative infinity?
How do I know that the limit is infinite at all? (Why not just "does not exist")

Thanks

You have $\lim_{x \rightarrow -3^+} \frac{x+2}{x+3}$.

$\frac{x+2}{x+3} = 1 - \frac{1}{x + 3}$ and it should be clear from a graph of $y = 1 - \frac{1}{x+3}$ that $\lim_{x \rightarrow -3^+} \left( 1 - \frac{1}{x+3} \right) = - \infty$.

This is different from $\lim_{x \rightarrow -3} \left( 1 - \frac{1}{x+3} \right)$, which does not exist because the left hand limit is not equal to the right hand limit.
• Aug 29th 2009, 08:10 PM
DBA
Thanks you for the answer. But my problem is it, to define the limit without a graph.
With a graph I can see in which direction the infinity goes, but without that, how can I see if it is negative or positive infinity?

Thanks