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Hi tim,, this is how i think the first three should be done:
Q1
(3^x + 3^2x)^1/x,, taking 3^x out, we get
3*(1 + 3^x)^1/x,,,
Now limit of the above expression will be 3*e^[lim 3x*(1/x)]=3*e^3
x->infinity
Q2
Solving the expression we get,,
log[(1-x^r)/(1-x)] divided by (e^x -1)
log(1-x^r)-log(1-x) divided by (e^x -1)
,,, since they are of 0/0 form,,, we can apply L'hopital rule (diffentiating the expression),,,which will give us
-r*x^(r-1)/(1-x^r) - [-1/(1-x)] divided by e^x,,,
As x approaches to 0,, this expression will approach to 1,,
Q3:
For Domain ,,l [lxl]-2l should be greater than 3,,
which means [ lxl]-2 < -3 or [lxl]-2 >3,,, former is not a possibility so only [lxl] > 5,,so x should be either less than -6 or greater than 6,, so ithink it should be option b
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