Question

I don't understand the concept for Infinite Limits.

Answer 1

For infinite limits, I know that there are as lim x goes to a from the right = +or - infinity

Answer 2

but then i dont get the one sided limits and determining if it is a negative or positive infinity

Answer 3

This is from my book:

If the values of f(x) increase without bound as x approaches positive infinity or as x approaches negative infinity, then:
limit as x apporaches positive infinity of f(x)= positive infinity
AND limit as x approaches negative infinity of f(x) = positive infinity

Answer 4

and so i really dont get this.

Answer 5

A lot of the time, if you dont know the graph, you have to actually test points. For example, let's say we have

\[\lim_{x \rightarrow 2^{+}}\frac{ x ^{2}+3 }{ x-2 }\]

Now we may not immediately know what this graph will do as it approaches the asymptote, but thats fine. Knowing that we are approaching 2 from the right, I would pick two points on the right of 2. You can then see what happens to the y-values. DO they shoot up, down, or barely move. So let's try x = 3 first:

\[\frac{ (3)^{2}+3 }{ 3-2 }= 12\]
Now let's try 5/2 and see if our graph shoots up even more as to where we can say its clearly going to infinity:

\[\frac{ (\frac{ 5 }{ 2 })^{2}+3 }{ \frac{ 5 }{ 2 }-2 }= \frac{ \frac{ 37 }{ 4 } }{ \frac{ 1 }{ 2 } }=18.5 \]Because this was a pretty big jump, I would say its safe to say this goes to positive infinity. Kinda see what I did?

Answer 6

OH. so its safer to just plug in points and see if it goes up dramatically or goes down dramatically, cause those points u used were (3,12) and (5/2, 18.5) right?

Answer 7

wait, isnt 5/2 less than 3, how come the point on the y is higher than 3,12

Answer 8

It is less than 3, butits CLOSER to 2, meaning we're approaching it from the right like we want.

Answer 9

now im confused again
cause if its like what u said