1. ## Limits

Hi. I need help in this problem. If someone could help that will be very nice. Sorry but I don't know how to write math equations on a forum post.
Any help would be nice. Thanks

1. Knowing that f and g are two functions and:

lim f(x)= +oo (+infinite) and lim g(x)=0
x->a x->a

(a) Show that choosing conveniently the functions f and g and defining h=f*g we can have the follow situations:

(i) lim h(x) = +oo
x->a

(ii) lim h(x) = -oo
x->a

(iii) lim h(x) = 0
x->a

(iv) lim h(x) = 5
x->a

(v) lim h(x) = -5
x->a

(vi) lim h(x) do not exist , without being +oo or -oo
x->a

Note: The point a € R is a generic point.
€=belong

2. Originally Posted by usual_suspect
Hi. I need help in this problem. If someone could help that will be very nice. Sorry but I don't know how to write math equations on a forum post.
Any help would be nice. Thanks

1. Knowing that f and g are two functions and:

lim f(x)= +oo (+infinite) and lim g(x)=0
x->a x->a

(a) Show that choosing conveniently the functions f and g and defining h=f*g we can have the follow situations:
just to clarify:

$\lim_{x \to a}f(x) = \infty$ and $\lim_{x \to a}g(x) = 0$

and $h(x) = f(x) \cdot g(x)$? as opposed to $h(x) = f \circ g(x)$?

3. Originally Posted by usual_suspect
Hi. I need help in this problem. If someone could help that will be very nice. Sorry but I don't know how to write math equations on a forum post.
Any help would be nice. Thanks

1. Knowing that f and g are two functions and:

lim f(x)= +oo (+infinite) and lim g(x)=0
x->a x->a

(a) Show that choosing conveniently the functions f and g and defining h=f*g we can have the follow situations:
assuming you actually meant $h(x) = f(x) \cdot g(x)$

(i) lim h(x) = +oo
x->a
$f(x) = x^2$, $g(x) = \frac 1x$ and $a = \infty$

(ii) lim h(x) = -oo
x->a
$f(x) = -x^3$, $g(x) = \frac 1x$ and $a = - \infty$

(iii) lim h(x) = 0
x->a
$f(x) = \frac 1x$, $g(x) = x^2$ and $a = 0$

(iv) lim h(x) = 5
x->a
$f(x) = \frac 5x$, $g(x) = x$ and $a = 0$

(v) lim h(x) = -5
x->a
$f(x) = - \frac 5x$, $g(x) = x$ and $a = 0$

(vi) lim h(x) do not exist , without being +oo or -oo
x->a
$f(x) = e^x$, $g(x) = \frac {\sin x}{e^x}$ and $a = \infty$

(by the way, we can choose a right? if you're not allowed to, i leave it to you to modify my responses)

4. Originally Posted by Jhevon
just to clarify:

$\lim_{x \to a}f(x) = \infty$ and $\lim_{x \to a}g(x) = 0$

and $h(x) = f(x) \cdot g(x)$? as opposed to $h(x) = f \circ g(x)$?

Im suposing $h(x) = f(x) \cdot g(x)$

5. Originally Posted by Jhevon
assuming you actually meant $h(x) = f(x) \cdot g(x)$

$f(x) = x^2$, $g(x) = \frac 1x$ and $a = \infty$

$f(x) = -x^3$, $g(x) = \frac 1x$ and $a = - \infty$

$f(x) = \frac 1x$, $g(x) = x^2$ and $a = 0$

$f(x) = \frac 5x$, $g(x) = x$ and $a = 0$

$f(x) = - \frac 5x$, $g(x) = x$ and $a = 0$

i have to think of this one some more. i want a nice example

(by the way, we can choose a right? if you're not allowed to, i leave it to you to modify my responses)
Thanks Jhevon. I still don't understant how you get to the results. Can you explain a litle more.

6. Originally Posted by usual_suspect
Thanks Jhevon. I still don't understant how you get to the results. Can you explain a litle more.
it's kind of hard to explain, you just have to "see" it

for instance, the first. i wanted $\lim_{x \to a}f(x) = \infty$, $\lim_{x \to a}g(x) = 0$

since i want to get $\lim_{x \to a}h(x) = \infty$ i said to myself. how can i get the limit to go to infinty? well, i can "divide by zero" or i can have the function be, say x, or e^x or whatever such that when x goes to infinity, the function goes to infinity, so i chose the latter.

so ok, obviously now i want $\lim_{x \to \infty}$, so i choose $a = \infty$

now i need to get $f(x)$ and $g(x)$

well, i know that $\lim_{x \to \infty} \frac 1x = 0$ because i have seen that limit so many times (and you should have too). so that's my $g(x)$, now, what do i want $f(x)$ to be?

i know $\lim_{x \to \infty}x = \infty$, so could that be $f(x)$? well, no, since i would have $h(x) = x \cdot \frac 1x = 1$ which does not go to infinity as x goes to infinity. so i realize, i just need one more factor of x to get my $\lim_{x \to \infty} x = \infty$, so i made $f(x) = x^2$. so now, $\lim_{x \to \infty}f(x) = \infty$, $\lim_{x \to \infty}g(x) = 0$ and $\lim_{x \to \infty}h(x) = \lim_{x \to \infty}x^2 \cdot \frac 1x = \lim_{x \to \infty}x = \infty$ as desired.

similar reasoning went into the others. just try to think of well-known limits and tweek them to fulfill the conditions. as you see, i did not use any complicated functions (well, the last one was kind of complicated, but only slightly), just keep it simple.

7. Originally Posted by Jhevon
it's kind of hard to explain, you just have to "see" it

for instance, the first. i wanted $\lim_{x \to a}f(x) = \infty$, $\lim_{x \to a}g(x) = 0$

since i want to get $\lim_{x \to a}h(x) = \infty$ i said to myself. how can i get the limit to go to infinty? well, i can "divide by zero" or i can have the function be, say x, or e^x or whatever such that when x goes to infinity, the function goes to infinity, so i chose the latter.

so ok, obviously now i want $\lim_{x \to \infty}$, so i choose $a = \infty$

now i need to get $f(x)$ and $g(x)$

well, i know that $\lim_{x \to \infty} \frac 1x = 0$ because i have seen that limit so many times (and you should have too). so that's my $g(x)$, now, what do i want $f(x)$ to be?

i know $\lim_{x \to \infty}x = \infty$, so could that be $f(x)$? well, no, since i would have $h(x) = x \cdot \frac 1x = 1$ which does not go to infinity as x goes to infinity. so i realize, i just need one more factor of x to get my $\lim_{x \to \infty} x = \infty$, so i made $f(x) = x^2$. so now, $\lim_{x \to \infty}f(x) = \infty$, $\lim_{x \to \infty}g(x) = 0$ and $\lim_{x \to \infty}h(x) = \lim_{x \to \infty}x^2 \cdot \frac 1x = \lim_{x \to \infty}x = \infty$ as desired.

similar reasoning went into the others. just try to think of well-known limits and tweek them to fulfill the conditions. as you see, i did not use any complicated functions (well, the last one was kind of complicated, but only slightly), just keep it simple.
ok.Thanks again. That will help very much.