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November 15th, 2014, 04:12 AM   #1
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How to approach these type of questions?

Hello, I need to prove or disprove certain claims. I will give you 2 as examples (I did manage to solve some others but just with luck)

Prove or disprove with counter-example the following:

1)lim(x)->infinity f(x)/x = infinity ===> lim(x)->infinity (f(x)-x)=infinity
2)lim(x)->infinity f(x)/x = 1 ===> lim(x)->infinity (f(x)-x)=0

How to start? what am I looking for to prove or disprove? And by the way what I managed to solve so far was only disproving with an example. How am I supposed to prove anyway if a claim is right?

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Last edited by noobinmath; November 15th, 2014 at 04:15 AM.
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November 15th, 2014, 04:38 AM   #2
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Try looking at the limit of $$\frac{\frac{f(x) - x}{x}}{\frac1x}$$or$$\frac{x(f(x) - x)}{x}$$
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Last edited by v8archie; November 15th, 2014 at 05:31 AM.
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November 15th, 2014, 05:03 AM   #3
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Can you explain? and does it match the (1) or (2) claim?
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November 15th, 2014, 05:03 AM   #4
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Here's an alternative way to look at 1):

You know that f(x)/x becomes unbounded as x gets large. So, this means that f(x) gets greater and greater as a mutliple of x.

So, if f(x) is getting large and x is getting large, then f(x) - x must be getting large.

But, maybe there's another way that f(x)/x -> infinity? Maybe if f(x) and x are both getting small. What then?
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November 15th, 2014, 05:26 AM   #5
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Quote:
Originally Posted by noobinmath View Post
Can you explain? and does it match the (1) or (2) claim?
One or the other will probably work for each.

$$\lim_{x \to \infty} \left(f(x) - x\right) = \lim_{x \to \infty} \frac{x}{x}\left(f(x) - x\right) = \lim_{x \to \infty} x\left(\frac{f(x)}{x} - \frac{x}{x}\right) = \lim_{x \to \infty} x\left(\frac{f(x)}{x} - 1\right)$$
and both $x$ and $\frac{f(x)}{x}-1$ go to infinity.

For the second, try thinking about $f(x) = \sqrt{x^2 + bx +c}$ and consider $\left(f(x)-x\right)\frac{f(x)+x}{f(x)+x}$.
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Last edited by v8archie; November 15th, 2014 at 05:51 AM.
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November 15th, 2014, 06:04 AM   #6
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Quote:
Originally Posted by Pero View Post
But, maybe there's another way that f(x)/x -> infinity? Maybe if f(x) and x are both getting small. What then?
How can it be that both are getting small ? x goes to infinity.

Quote:
Originally Posted by v8archie View Post
One or the other will probably work for each.

$$\lim_{x \to \infty} \left(f(x) - x\right) = \lim_{x \to \infty} \frac{x}{x}\left(f(x) - x\right) = \lim_{x \to \infty} x\left(\frac{f(x)}{x} - \frac{x}{x}\right) = \lim_{x \to \infty} x\left(\frac{f(x)}{x} - 1\right)$$
and both $x$ and $\frac{f(x)}{x}-1$ go to infinity.

For the second, try thinking about $f(x) = \sqrt{x^2 + bx +c}$ and consider $\left(f(x)-x\right)\frac{f(x)+x}{f(x)+x}$.
I don't understand. in the 1st claim I am supposed to show that if
lim(x)->infinity f(x)/x = infinity then lim(x) ->infinity (f(x)-x) = infinity not the other way around?

and about "For the second, try thinking about ..." I don't understand how do you get to these decisions, how do you get to this decision to do that? I am really confused

I really tried what you said about the second one and took f(x)=sqrt(x^2+3x+1) and when it's divided by x then the limit at inifinity is 1, and when you do f(x)-x it's -3/2. But how did you think about that? You're good.

Last edited by noobinmath; November 15th, 2014 at 06:51 AM.
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November 15th, 2014, 08:03 AM   #7
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[QUOTE=noobinmath;212755]How can it be that both are getting small ? x goes to infinity.

Yes, bit of a brain malfunction. I must have been thinking f(x)/g(x).

It made you think about it anyway!
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November 15th, 2014, 08:29 AM   #8
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I got to the second one because of experience telling me what would work.

But also, if you follow the process used for the first one you get to the limit of $\infty \cdot 0$ which is undefined and can be any value at all. Then the trick is to find an example where the limit is non-zero. Since we have $\frac{f(x)}{x} \to 1$, we know that the highest power of $x$ must be zero, but for it to be bigger than $x$, we must have some other term with a smaller power of $x$. Actually, thinking about it $f(x) = x + c$ works too, and quite trivially so.

For the first, we showed that $$\lim_{x \to \infty} f(x) - x = \lim_{x \to \infty} x \left(\frac{f(x)}{x} - 1\right)$$Then, using the limit we were given we find that the limit must be $+\infty$. This is the required proof.
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November 15th, 2014, 01:17 PM   #9
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v8archie you're good!

Just something about the second one that I still don't understand (Probably because of my English) :

"we know that the highest power of x must be zero, but for it to be bigger than x, we must have some other term with a smaller power of x"

In the example x+c, which is working as well , the power of x must be zero because x^0=1? Is that what you meant? but x+c the power is 1 and not 0.
Also you said we must have other term with a smaller power of x. and again, in the example x+c, when you do (x+c)/x , both have the same power, and non have smaller of the other

I'm new to calculus and it's really confusing so thanks for your help and patience
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November 15th, 2014, 04:46 PM   #10
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Quote:
we know that the highest power of x must be one, but for it to be bigger than x, we must have some other term with a smaller power of x
I apologise, there was a typing error there.

The highest power of $x$ in $f(x)$ must be 1 to get $\frac{f(x)}{x} \to 1$. But we may have smaller powers of $x$ too - and a constant $c$ is a term in $x^0$.

These smaller powers disappear in the limit when we divide by $x$, but not when we subtract $x$.
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