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Calculus/L'Hospital's Rule
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Question
Find the following limits:
a. lim x->Infinity ((lnx)^3)/x^2
b. lim x->0 (cosx)^(1/x^2)
a. applying L'Hospital rule we get :
3(lnx)^2*(1/x) (6lnx)
------------- = ---------
2x 2x^2
Then we applyit another time , we get :
6/x 3
---- = -----
4x 2x^2
& thisform goes to zero when x goes to infinity.
b. Let's set y=(cosx)^(1/x^2) then Lny=(1/x^2)Ln(cosx).
Now we need to know the limit of (1/x^2)Ln(cosx). In order to
do so , we will use L'Hospital's Rule :
-(1/cosx)*sin(x)
--------------
2x
Applying second time we get :
(1/cosx)^2
----------
2
& this form goes to -0.5 when x goes to zero.
So Lny=-0.5 -> y=1/sqrt(e) , & that means that our originional limit goes
to 1.
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