find the limit as x --> 0+ of (1-3x)^(1/x)

Question

how would I know that I'm supposed to use 'ln' in this question? and why is the final answer e^(-3) instead of just -3?

Answer 1

First of all, this looks similar to the limit definition of e, which is

e = lim_n->∞(1+1/n)ⁿ

Let 1/n = -3x

x = -1/(3n)

1/x = -3n

As x->0, n ->∞

Substituting into the original equation:

lim_n->∞(1+1/n)^-3n = [lim_n->∞(1+1/n)ⁿ]^-3 = [e]^-3 = e^-3

^

Answer 2

y = (1 - 3x)^1/x; lny = ln(1 - 3x)/x; lim x→0⁺ln(1 - 3x)/x = lim x →0⁺ [-3/(1 - 3x)]/1 = - 3.

So, lim x→0⁺(1 - 3x)^1/x = e^-3

^{We here used L'opital rule.}