lim x ->0 (1-3x)^1/x

Question

lim x ->0 (1-3x)^1/x

The other question is:

∫ 0 to infinity te^-5t dt

Answer 1

1) The trick is to make the exponent come down by finding the limit of Y = ln(y) = 1/x *ln(1-3x)

This now qualifies for l'Hopital's rule Lim as x -->0 of ln(1+3x)/x (0/0) which leads to

Lim as x --> -3/(1-3x)/1 after taking the derivative of num and denom. Lim Y = -3

Since Y = -3, y = e^Y = e^-3

2) Use Integration by parts. I remember LITPET (picture cat getting high) for what term should be differentiated (logs, inverse trig, power, exp, trig). In this case u = t and e^-5t is v

If you do this you should get te^-5t/(-5) - e^-5t/25 evaluated from 0 to inf. The inf limits are 0 (e^-5t goes to 0 and much faster than any power function) and that leaves -(-1/25) = 1/25 or .04