# Two problems

1. Jun 2, 2003

### PrudensOptimus

here's what i got:

(1 + (lim x ->inf, sinx/x))/(1 + (lim x->inf, cosx/x))

note the "1"s came from x/x from the original limit.

And lim x->inf, cosx/x becomes 0.
but I don't understand how lim x->inf, sinx/x becomes 0.
sinx/x = {1 - x^2/3! + x^4/5! - ... + ...} since x approaches infinity, that series would be 1 - inf + inf - inf, eventually it is still infinity.

so, we have (1 + inf)/(1 +0) = inf, but the text says it approaches 1, so it much be something wrong i did with the sinx/x thing.

2. Jun 2, 2003

### PrudensOptimus

and another question about lim x->+-inf, (x+1)/12: how does it become +-inf? isn't it suppose to be No LImit because (x+1)/12 as x -> inf becomes 1+0/0??

3. Jun 2, 2003

### suffian

The limit is Sin[x]/x is more easily understood as zero if you apply the sandwich thereom.

-1 <= Sin[x] <= 1
-1/x <= Sin[x]/x <= 1/x, (x > 0)

Limit[ x->inf, -1/x ] = 0
Limit[ x->inf, 1/x ] = 0

So the function Sin[x]/x is always "squeezed" inbetween the functions 1/x and -1/x for x>0 by the second statement. Since these two functions themselves approach zero from above and below, Sin[x]/x, which is always between the two, is also forced go to zero.

4. Jun 2, 2003

### suffian

Well, by saying that the limit of a function as x approachs c is infinity really means: "this limit doesn't exist, but the closer you get to the number c, the larger the value of f(x) will get". It's a conveniant way of describing how the function misbehaves near the point c or towards infinity (or minus infinity).

5. Jun 2, 2003

### Hurkyl

Staff Emeritus
This kind of series is an "indeterminate form", meaning the only thing it tells us is that we have to do more or different work.

The easiest way to see how the answer comes is to think back to trig and remember that (sin x) has a maximum value.

6. Jun 2, 2003

### NEOclassic

because sin x isn't necessarily inf as x -> inf and the reciprocal of 1/x with x -> inf is naturally x -> zero why not check the limit of the reciprocal which gives x/sin x, as x -> zero, which has a limit of one.
Cheers, Jim

7. Jun 3, 2003

### plus

To see the answer look at what happens when x gets large, and the difference between x and sin(x).

e.g. when x=100000000, |sin(x)|<= 1

8. Jun 3, 2003

### KLscilevothma

Another good way to find the limit value of sinx/x as x tends to inf is to use the following property.
Let f(x) and g(x) be two functions defined on an interval containing a, possible except a.
If f(x) is bounded and lim x->a h(x) = 0,
then lim x->a h(x)f(x)=0

lim sinx / x
x->[oo]

notice that 1/x tends to zero while |sinx|<=1 which is bounded, as x tends to inf.

Therefore the limit tends to zero.