calculus
posted by Jake .
Lim sin2h sin3h / h^2
h>0
how would you do this ?? i got 6 as the answer, just want to make sure it's right.
and i couldn't get this one (use theorem 2)
lim tanx/x
x>0
and also this one (use squeeze theorem to evaluate the limit)
lim (x1)sin Pi/x1
x>1

the first one is correct.
Here is a simple way to check your limit answers if you have a calculator
pick a value very "close" to your approach value, in this case I would pick x = .001
evaluate using that value, (you are not yet dividing by zero, but close)
I got a value of 5.999985 which I would say is close to 6
for lim tanx / x as x >0
= lim (sinx/cosx)/x
= lim (sinx/x)*lim 1/cosx as x > 0
= 1*1 = 1
If you meant lim (x1)sinπ/(x1)
wouldn't the last one simply be sinπ or 0 ?
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