Find the positive integers k for which
lim ->0 sin(sin(x))/x^k
exists, and then find the value the limit.
(hint:consider first k=0, then k=1. Find the limit in these simple cases. Next take k=2 and finally consder k>2 and find the limit in these cases as well)
I did this for the first one
lim x->0 sin(sinx))/x^0=0
then idk how to do the other ones
k=1 limit sinx/x=1
k=2 limit sinx/x *1/x= undefined
and so on.
for k=2 lim x->0 sin(sin(x))/x^2= error
so undefined So do I have to do sided limits? how?
To find the positive integers k for which the limit of (sin(sin(x))/x^k) exists, we need to consider each case separately.
1. For k = 0:
To evaluate the limit when k = 0, we have:
lim x->0 sin(sin(x))/x^0
Since x^0 is always 1, the expression simplifies to:
lim x->0 sin(sin(x))
This limit evaluates to 0 because sin(x) oscillates between -1 and 1, and sin(sin(x)) will also oscillate between -1 and 1. As x approaches 0, sin(sin(x)) approaches 0.
2. For k = 1:
To evaluate the limit when k = 1, we have:
lim x->0 sin(sin(x))/x^1
Since x^1 is just x, the expression becomes:
lim x->0 sin(sin(x))/x
This limit evaluates to 1. One way to see this is by using L'Hopital's rule and taking the derivative of both the numerator and denominator with respect to x.
3. For k = 2:
To evaluate the limit when k = 2, we have:
lim x->0 sin(sin(x))/x^2
Again, using L'Hopital's rule, we take the derivative of both the numerator and denominator with respect to x, which gives:
lim x->0 (cos(x) * cos(sin(x)))/2x
This limit can be evaluated to 1/2. Another way to see this is to approximate sin(x) by x for small x, and then use a Taylor series expansion.
4. For k > 2:
For values of k greater than 2, as x approaches 0, the denominator x^k approaches 0 faster than sin(sin(x)), resulting in a limit of infinity. Thus, the limit does not exist for k > 2.
In summary:
- The limit exists and is 0 for k = 0.
- The limit exists and is 1 for k = 1.
- The limit exists and is 1/2 for k = 2.
- The limit does not exist for k > 2.