# calculus

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 (x-1)sin Pi/x-1
x-->1

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 (x-1)sinπ/(x-1)
wouldn't the last one simply be sinπ or 0 ?

posted by Reiny

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