could someone explain how you would use 'ln' to solve this? lim x--> 0+ for (4x)^(5x)

Question

and since it's 0 from the right side, would I still plug in 0? or a small number like 0.001?

Answer 1

let y be ln(4x)^5x ). Find the limit of y as x goes to 0

y = (5x)ln(4x) (This is not a form that we can use L'Hopital's rule (It's 0*(-inf) as x goes to 0)

rearranging:

y = 5ln(4x)/(1/x) (Now you can take derivative of numerator and denominator in order to find the limit

Lim of y as x goes to 0 = 20(1/4x)/(-1/x²) = 0

If y goes to 0 then (4x^5x) must go to 1. Of course, if you keep substituting smaller and smaller numbers for x, you will see that 1 is being approached.

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