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Calculus Question help
Started by FightingIrish01
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this is off topic with this thread. but still, i'm surprised i remember calculus from way back more than 2 years ago.
the answer is zero b/c the numerator is limited to a specific small number while the denominator becomes something like "a million times a million is a very big number". so a very small number over a very big number looks like a zero.
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insideedition is right
think about it, cos(x) for all x is between -1 to 1, so the numerator can only be -2 to 2, the denominator simply goes to infinity. so the limit is 0.
Wait, but I though limit as x-> infinity of sin(x) is undefined because that is an oscillating function. Thus, what about limity x->infinity of cos(x)
Thanks.
Thanks.
what?!
okay here's the deal
x/x^2 -> 0 as x->infinity, this point is explained pretty well by inside_edition
now 2cos (0) is 2, which is the answer to this question.
this can also be solved using a TI 89 calculator by it's limit() function.
Dang, my high school cal teacher must drilled these concept in me hard, i still solve calc like nobody's business after all these years.
okay here's the deal
x/x^2 -> 0 as x->infinity, this point is explained pretty well by inside_edition
now 2cos (0) is 2, which is the answer to this question.
this can also be solved using a TI 89 calculator by it's limit() function.
Dang, my high school cal teacher must drilled these concept in me hard, i still solve calc like nobody's business after all these years.
FightingIrish,
I hope you're not from ND, but I'll answer your question anyways 🙂. It's true that limit as x-> infinity of sin and cos are undefined b/c they are oscillating functions. But as previously said, when in the numerator over x^2 the possible range of sin or cos (-1 to 1) becomes negligible when compared to the overall value after dividing by x^2.
I hope you're not from ND, but I'll answer your question anyways 🙂. It's true that limit as x-> infinity of sin and cos are undefined b/c they are oscillating functions. But as previously said, when in the numerator over x^2 the possible range of sin or cos (-1 to 1) becomes negligible when compared to the overall value after dividing by x^2.
but when you put in "divided by x^2" under those specific conditions, does it make a difference whether the answer is -.000000000001 or +.000000000001? it's true that those functions alone will lead to undefined functions b/c they oscillate. it is similar to the fact that you know that your pencil is in a specific location even though the molecules in your pencil move around and oscillate. in other words, the oscillation is very small compared to the whole picture.
as many overs have stated in various ways...
a limit for an oscillating function is not defined if the oscillation never approaches any final value.
however this function will always oscillate, but the amplitudes of each successive peak or valley will be nearer to 0, thus as x goes to infinity, the function goes to 0.
i miss my calc days...
a limit for an oscillating function is not defined if the oscillation never approaches any final value.
however this function will always oscillate, but the amplitudes of each successive peak or valley will be nearer to 0, thus as x goes to infinity, the function goes to 0.
i miss my calc days...
Ok, sorry to add on to this...
What is the answer to:
1.) Lim x->0 of cos(x)/x
2.) Lim x-> of cos(x)/x^2
I think I am set on this. I am sorry if this sounded simple to you guys.
What is the answer to:
1.) Lim x->0 of cos(x)/x
2.) Lim x-> of cos(x)/x^2
I think I am set on this. I am sorry if this sounded simple to you guys.
we aren't getting across to you so we'll try it a different way
1. lim x->0 of cos(x)/x == (lim x->0 cos x) * (lim x->0 1/x)
2. lim x-> of cos(x)/x^2 == (lim x->0 cos x) * (lim x->0 1/x^2)
as x-> 0, cos x goes to 1
as x->0, 1/x goes to infinity
as x->0, 1/x^2 goes to infinity also
so.........
if you were to multiply these together:
for 1: (1) * (infinity) = infinity
for 2: (1) * infinity = infinity
make sense?
Thanks. I understood your point but I guess I though you could not break up functions like you did because if so then by original question and answer would be valid because limx->infinity of cos(x) is Does not exist.
ok, you make a point but there is an explanation for this as well
although the lim x->inf cos(x) is DNE, it is in fact definable to be somewhere between -1 and 1 which when divided by ever increasing large numbers gets closer and closer to 0
so in this one case
DNE * 1/infinity still equals 0 not DNE
