Thread: 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

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 and take the log of both sides:




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


Note that the limit of is 1, so you're left with
, whose limit is 0.

But remember that we just measured the limit of 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)

please tell the steps also

Originally Posted by ayush

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

please tell the steps also

You need to start a new thread. Please spend some time reading the forum rules.

Start by simplifying the expression



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

Originally Posted by ayush

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

please tell the steps also

Just go ahead and do the multiplication: . What happens to that as x goes to infinity?

You could also do this "piece by piece". What is the limit of as x goes to infinity? Now, what is the limit of that, divided by x, as x goes to infinity?