# Thread: help me in this kind of questions

1. ## help 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

2. Originally Posted by ayush
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 $y = x^{sin2x}$ and take the log of both sides:

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

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 $\frac{\frac{1}{x}}{-2\frac{cos(2x)}{sin^{2}(2x)}}$
$\frac{sin^{2}(2x)}{-2xcos(2x)}$

Note that the limit of $\frac{sin(2x)}{2x}$ is 1, so you're left with
$\frac{sin(2x)}{-cos(2x)}$, whose limit is 0.

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

Good luck!!

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

4. Originally Posted by ayush
lim
x->infinity ((1+1/x^2)/x)

Start by simplifying the expression

$\frac{\frac{x^2+1}{x^2}}{x} = \frac{x^2+1}{x^3}$

as x blows up the ratio goes to zero because x^3 grows faster.

5. Originally Posted by ayush
lim
x->infinity ((1+1/x^2)/x)

Just go ahead and do the multiplication: $(1+ \frac{1}{x})\frac{1}{x}= \frac{1}{x}+ \frac{1}{x^2}$. What happens to that as x goes to infinity?
You could also do this "piece by piece". What is the limit of $1+ \frac{1}{x}$ as x goes to infinity? Now, what is the limit of that, divided by x, as x goes to infinity?