limits help?

Question

evaluate
Lim { (sinx)^1/x + (1/x)^sinx }
x->0

Answer 1

You need to specify right or left limits: the left limit does not exist (a negative number to an irrational exponent) but for the right hand limit x -> 0+ you use the logarithm trick

(1/x)^sin(x) = exp(log( (1/x)^sin(x) ) = exp( -sin(x) * log(x) )

Then use L'Hopital's on -sin(x)*log(x) = -log(x)/ [1/sin(x)] to get (-1/x)/ [-1/sin(x)^2*cos(x)] = sin(x)^2/ (x*cos(x)) -> 0 as x ->0+

The other term sinx^(1/x) is not indeterminate: 0^infinity definitely -> 0

So then you end up with 0+exp(0) = 1

[Don't forget to put the limit 0 BACK in the exponent!]

Answer 2

If x is greater than zero, then

Lim { (sinx)^1/x + (1/x)^sinx }
x->0
=Lim { e^[ln(sinx)]/x + e^[ln(1/x)]/(1/sinx) }
x->0
Lim { e^(cosx/sinx) + e^[sin^2x/(xcosx)}
x->0
= infinity + 1
= infinity

If x is less than zero, then let y = -x. Now y is greater than zero. Therefore, we have
Lim { (sinx)^1/x + (1/x)^sinx }
x->0
=Lim { (-siny)^(-1/y) + (-1/y)^(-siny) }
y->0

Since the base is negative, the limit cannot exist.