Solution to Find Limit as x Approaches -5 from Left of 3x/2x+10

[sqidwarf]

1. limit as x approaches -5 from the left
of 3x/2x+10

Homework Equations

The Attempt at a Solution

I know the solution is infinity but I don't know how to prove it. Should I use the formal definition or is there another way?

 

[lanedance]

simply writing it as:
\frac{3}{2} \frac{x}{(x+5)}
should be sufficient to show the numerator goes to -5*2, whilst denominator is goes to 2*0, so the answer isn't positive infinity...

 

[lanedance]

otherwise fromally, you could show for any poistive integer N, you can choose d such that for
-5-x < d then f(x) < -N

though bit of overkill i think

 

[Elucidus]

You can use the following theorems:

For any real number a and positive integer m:

\lim_{x \rightarrow a^+} \frac{1}{(x-a)^m} = \infty

\lim_{x \rightarrow a^-} \frac{1}{(x-a)^m} = \left\{ \begin{array}{rl}<br /> \infty, &amp; \text{if }m \text{ is even} \\<br /> -\infty, &amp; \text{if }m \text{ is odd}<br /> \end{array}

If \lim_{x \rightarrow a}f(x) = \infty and \lim_{x \rightarrow a}g(x) = L &gt; 0 \text{ or } \infty then

\lim_{x \rightarrow a}f(x) \cdot g(x) = \infty

(Note this last one works if approaching a from either the left or the right too.)

These theorems are provable with delta-epsilon proofs.

In you exercise let g(x) = 3x/2 and f(x) = 1/(x+5).

I hope this helps.

--Elucidus