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
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!!
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?