please solve this question and also show the steps of solution

1) Lim x->0 x^(2sinx)

ie lim x tends to 0, x to the power 2sinx

2) is L-hospitals rule valid for 0 to the power 0 condition

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- Jun 21st 2009, 10:14 AMayushhelp me in this kind of questions
please solve this question and also show the steps of solution

1) Lim x->0 x^(2sinx)

ie lim x tends to 0, x to the power 2sinx

2) is L-hospitals rule valid for 0 to the power 0 condition - Jun 21st 2009, 10:30 AMapcalculus
Set $\displaystyle y = x^{sin2x}$ and take the log of both sides:

$\displaystyle ln y = sin(2x) ln(x)$

$\displaystyle ln y = \frac{ln(x)}{csc(2x)}$

The limit of the right hand side as x goes to zero can be found using L'Hopital's Rule. (verify conditions)

So the limit as x tends to 0 of $\displaystyle \frac{\frac{1}{x}}{-2\frac{cos(2x)}{sin^{2}(2x)}}$

$\displaystyle \frac{sin^{2}(2x)}{-2xcos(2x)}$

Note that the limit of $\displaystyle \frac{sin(2x)}{2x}$ is 1, so you're left with

$\displaystyle \frac{sin(2x)}{-cos(2x)}$, whose limit is 0.

But remember that we just measured the limit of $\displaystyle ln y$ equal to 0, which means the limit of y, your original expression, is 1. Verify numerically and graphically.

Good luck!! - Jun 21st 2009, 10:59 AMayushplease help me in this question also
lim

x->infinity ((1+1/x^2)/x)

please tell the steps also - Jun 21st 2009, 11:16 AMapcalculus
- Jun 21st 2009, 11:18 AMHallsofIvy
Just go ahead and do the multiplication: $\displaystyle (1+ \frac{1}{x})\frac{1}{x}= \frac{1}{x}+ \frac{1}{x^2}$. What happens to that as x goes to infinity?

You could also do this "piece by piece". What is the limit of $\displaystyle 1+ \frac{1}{x}$ as x goes to infinity? Now, what is the limit of**that**, divided by x, as x goes to infinity?