Question

what is lim x->0 (1+3x)^(1/x)

Answer 1

Rather tricky, do you know about L'Hospitals Rule?

Answer 2

yes but I got stuck around ln(1+3x)/ x^-1

Answer 3

Should be ln(1+3x)/x

Answer 4

Then when x approaches 0 you get 0/0, so you can use L'Hospitals Rule.

Answer 5

because ln(1)=0

Answer 6

I don"t think I did it right but is the answer 3x^2/ 1+3x

Answer 7

that's not what I got, let me show you how I thought about this problem, but as I said before, it's tricky.

The first thing you want to do is to get rid of that exponent, you have already done that it seems I used the following substitution

\[ \large e^u = y = \ln(3x+1)^{\frac{1}{x}} \\
\therefore u = \frac{\ln(3x+1)}{x} \]

now, since this substitution didn't change much we can let the same function approach zero

\[ \lim_{u \to 0} = \frac{0}{0} \]

next you take the derivative of the numerator and the denominator, individually, not the quotient rule, you will get

\[ \large \frac{3}{3x+1}\] this limit as x approaches zero goes to 3, so u=3

and your final limit is

\[ \large e^3\]

Answer 8

In the first line I made a mistake, there is no natural log there yet, that's what I intended to do in the second step to get rid of the exponent.

Answer 9

that makes so much more sense thanks.

Answer 10

you are welcome