calculus limit question?

Question

lim k^2 * sin^2 ( 1 / k )
k->inf


How is it 1, not 0?

Answer 1

This is easier to follow if we substitute z = 1/k, getting limit sin^2(z) / z*2 as z goes to 0. L'Hopital's rule is now used; we take derivatives above and below to get 2 sin(z) cos(z) / 2z. But this also goes to 0/0 at the limit, so we have to do it again; we'll cancel out the 2's, note that the derivative of the denominator is 1, and cheat by assuming cos(z) = 1 [one could use the product rule, but why bother if you don't have to? Exercise: explain why this cheat is valid.]. This leaves us with taking the derivative of sin(z), which is cos(z), which at 0 is 1. By the time the dust settles, we have 1/1, or 1.

Answer 2

For the above question, when k approaches infinite, the function becomes inf * 0, which is equilvalent to inf/inf. In this situation, the limit of the above function is said to be in an indeterminate form. To find the limit, we have to apply L'Hospital's rule.

Basically, we need to put the function in fraction form, then take the derivative of the top and bottom "individually" and evaluate the limit. If the limit is still inf / inf (or 0/0), we will have to apply the L'Hospital's rule again.

For this particular question,

lim k^2 * sin^2 (1/k) = inf ^2 * sin^2 (1/inf) = inf * 0 = inf / inf
k -> inf

Applying L' Hospital's rule:

First put it in fraction form:

lim [sin^2 (1/k) / k^-2]

= lim {[1/2 - 1/2(cos 2/k)] / [k^-2]} (applying double angle formula to top)

Then take the derivative of top and bottom individually:

Top: d/dx {[1/2 - 1/2(cos 2/k)] = 1/2 sin (2/k) * (-2/ k^2)

= - sin (2/k) / k^2

Bottom: d/dx (k^-2) = -2(k^ -3)

Top/Bottom: sin (2/k) / 2*k^-1

Evaluate this limit, we still got 0/0; therefore, apply L'Hospital's Rule again

Top: d/dx [sin (2/k)] = cos (2/k) * -2 / k^2

Bottom: d/dx [2*k^-1] = -2 * k^-2

Top/Bottom:

[-2*cos (2/k) / k^2] / (-2 * k ^-2)

= [-2*cos(2/k)] / [-2*k^2*k^-2]

= cos(2/k)

Evaluate this limit as k approaches infinite:

lim cos(2/k) = lim cos (2 / inf) = lim cos (0) = 1
k -> inf

Therefore,

lim k^2 * sin^2 (1/k) = 1
k -> inf

(Took me long time to type this up......)

Answer 3

Let k = (1/n).

As k -> inf , n -> 0

Thus the limit becomes:


lim sin^2 (n)
n -> 0 n^2

Thus by the formula it evalutes to 1.

Answer 4

lim k^2 * sin^2 ( 1 / k )
k->inf
This leads to the indeterminate form infinity*0.
So let x = 1/k and our limit becomes :
= lim (sin^2(x))/x^2
x--> o
Now it is form 0/0 so L'Hospital's Rule can be applied:
Take successive derivatives of the numerator and denominator separately and text the limit each time as x--> 0.

You will get sin(2x)/2x and the 2cos2x /2 where thelimit is 1.

Answer 5

Here's how I'd do that problem.
The maclaurin series for sin(x) is
sin(x) = x - x^3 / 6 + ...
sin^2 (x) = x^2 - x^4 / 3 + ...

so sin^2(1/k) = (1/k^2) - (1/3)*(1/k^4) + ...
k^2 * sin^2 (1/k) = 1 - (1/3)*(1/k^2) + ...
as k goes to inf, the other terms in the series go to zero, leaving the limit of 1.

Answer 6

lim sin(X)/X=1 provided x->0
here 1/k ->0 as k->inf...
so
=> lim k^2 * sin^2 ( 1 / k )
k->inf
=>lim{sin(1/k)/(1/k)}*{sin(1/k)/(1/k)}
k->inf
=>lim{1}*{1}
k->inf
=>lim =1
k->inf

and hence not zero...
hope it helped...