Limit of (x*sin(1/x))^x^2. x tends towards infinite.?

Question

the function is to the power x^2. i hope it is clear....

Answer 1

Do you think you'll learn anything from this? I don't think math majors just scour Yahoo Answers day and night looking to answer random math problems.

What if your friend gives you an answer? Do you contemplate HOW they came to that answer, or do you just copy and turn it in?

Study the book, ask those in your class for help, ask your teacher for help. Don't expect free lunches in life. In the math realm, from experience I've learned it's harder to examine someone else's solution to get the process. It's in the book, or your notes. Trust me.

Answer 2

Let y = (x sin(1/x))^(x²) Then

ln(y) = x² ln(x sin(1/x))

Under conditions that we will assume are met, the limit of the log equals the log of the limit. So we want to consider

lim x² ln(x sin(1/x))

x→∞

Another transformation is helpful here. Let z = 1/x. Then the above limit is equivalent to

lim ln(sin(z) / z) / z²

z→0+

Using L'Hopital's Rule, we get

lim {(1/(sin(z)/z)[z cos(z) - sin(z)]/z²}/2z

z→0+

which simplifies to

lim (z cos(z) - sin(z))/(2z²sin(z))

z→0+

Still 0/0, so we need to use the Rule again.

lim (cos(z) - z sin(z) - cos(z))/[2(2z sin(z) + z² cos(z))]

z→0+

which simplifies to

lim  (-sin(z))/[2(2 sin(z) + z cos(z))]

z→0+

One more time!

lim (-cos(z))/[2(2 cos(z) + cos(z) - z sin(z))]

z→0+

At last, we get the limit -1/6

This is the log of the original limit, so the original limit is e^(-1/6)