Cos(ax)/a ,when a->0?

  • #1
Hi all ,
I'm an electronics engineer and I'm trying to solve this two equations:

1.
sin(ax)/a, when a=0
I do it this way:
Multiply by x and make a limit transition
lim sin(ax)x/ax, as lim sin(ax)/ax = 1
ax->0 ax->0
I infer:
lim sin(ax)x/ax = x
ax->0

2.
Unfortunatelly I'm nowhere with
cos(ax)/a, when a=0

Any help :)
P.S.
Sorry for the bad english.
 

Answers and Replies

  • #2
CompuChip
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Are you allowed to use L'Hopital's rule?
Since your first limit is of the from
[tex]\lim_{a \to 0} f(a) / g(a)[/tex] where [tex]\lim_{a \to 0} f(a) = \lim_{a \to 0} g(a) = 0[/tex]
it would apply in this case.

The second one is easier, because it is of the from
[tex]\lim_{a \to 0} f(a) / g(a)[/tex] where [tex]\lim_{a \to 0} f(a)[/tex] exists and is finite, i.e. it is similar to
[tex]\lim_{a \to 0} f(0) / g(a)[/tex]
 
  • #3
HallsofIvy
Science Advisor
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Hi all ,
I'm an electronics engineer and I'm trying to solve this two equations:

1.
sin(ax)/a, when a=0
I do it this way:
Multiply by x and make a limit transition
lim sin(ax)x/ax, as lim sin(ax)/ax = 1
ax->0 ax->0
I infer:
lim sin(ax)x/ax = x
ax->0

That is, as you say, a limit. It is NOT the value of sin(ax)/a "when a= 0". That has no value.

2.
Unfortunatelly I'm nowhere with
cos(ax)/a, when a=0

Any help :)
P.S.
Sorry for the bad english.
That is because the limit does not exist. The denominator is going to 0 as the numerator goes to 1. As a gets closer and closer to 0, the absolute value gets larger and larger.
 
  • #4
Hi, thanks for your replyes!

Dear CompuChip,
I really haven't investigate if the N1 limit
excits at all, so I can apply the L'hopital rule.

"The second one is easier", ok, that encaurages me!

Dear HallsofIvy,
I have to admite that I'm making a limit transition,
so i'm searching for the limit, but not for the exact value.

Thanks once again!
 
  • #5
391
0
the cosine is BOUNDED it can not be greater than 1 ,however 1/a is UNBOUNDED so the limit is infinite

if you expand the cosine(ax) into a taylor series in (ax) there is no linear part on 'x' only a constant term 1 and a quadratic term so there is no cancellation of the linear term and the limit is infinite
 
  • #6
CompuChip
Science Advisor
Homework Helper
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47
I don't know if this is legal, but I just made it up:
You can multiply by sin(ax)/sin(ax) (the limit point a = 0 is not in the domain we are considering anyway) and write it as
[tex]\frac{\cos(ax)}{\sin(ax)} \cdot \frac{\sin(ax)}{a}[/tex]
Now if you take the limit, the second factor is a standard limit (which you have just proven) going to one, and the first factor is 1/tan(a x). As the reciprocal of a function continuous at a x = 0 this limit does not exist, and therefore neither does the original limit.

(Of course, this sort of hinges on the fact that you can calculate [tex]\lim_{x \to 0} f(x)[/tex] as [tex]\lim_{x \to 0} f(x) g(x)[/tex] when [itex]\lim_{x \to 0} g(x) = 1[/tex]).
 
  • #7
Dear CampuChip,
Your solution is very interesting.
According to my first post the
lim sin(ax)/x = x, when ax->0,
but not to 1.
So if we put this into your solution:
lim cos(ax)/x = lim [cos(ax).x/sin(ax)], when ax->0.
10x,
Eltimir
 
  • #8
CompuChip
Science Advisor
Homework Helper
4,302
47
Yes, sorry, my bad.
You showed that
[tex]\lim_{a \to 0} \frac{\sin ax}{a} = x[/tex]
I claimed that
[tex]\lim_{a \to 0} \frac{\sin a}{a} = 1[/tex]
(which is a standard limit) and mistakenly said that that is what you had proven.
 

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