Find the following limit please

lim (x->inf) (1+1/x+2/x^2+3/x^3)^x

Note that this limit takes the form 1^infinity. To find this limit, start out by letting:
L = lim (x-->infinity) (1 + 1/x + 2/x^2 + 3/x^2)^x.

Then, by taking the natural logarithm of both sides:
ln(L) = lim (x-->infinity) ln[(1 + 1/x + 2/x^2 + 3/x^2)^x]
= lim (x-->infinity) x*ln(1 + 1/x + 2/x^2 + 3/x^2), since ln(a^b) = b*ln(a)
= lim (x-->infinity) ln(1 + 1/x + 2/x^2 + 3/x^2)/(1/x).
(Note that we needed to write this as a fraction to use L'Hopital's Rule.)

By applying L'Hopital's Rule:
ln(L) = lim (x-->infinity) ln(1 + 1/x + 2/x^2 + 3/x^2)/(1/x)
= lim (x-->infinity) [(-1/x^2 - 4/x^3 - 9/x^4)/(1 + 1/x + 2/x^2 + 3/x^3)](-1/x^2)
= lim (x-->infinity) (1 + 4/x + 9/x^2)/(1 + 1/x + 2/x^2 + 3/x^3)
= (1 + 0 + 0)/(1 + 0 + 0 + 0)
= 1.

Therefore, ln(L) = 1 and:
L = lim (x-->infinity) (1 + 1/x + 2/x^2 + 3/x^2)^x = e.

I hope this helps!

L = lim (1 + 1/x + 2/x^2 + 3/x^3)^x
lnL = lim ln((1 + 1/x + 2/x^2 + 3/x^3)^x)
lnL = lim xln(1 + 1/x + 2/x^2 + 3/x^3)
lnL = lim ln(1 + 1/x + 2/x^2 + 3/x^3)/(1/x)
Now you can use L'Hopital's rule because the numerator and denominator approach 0.
lnL = lim ((-1/x^2 - 4/x^3 - 9/x^4)/(1 + 1/x + 2/x^2 + 3/x^3))/(-1/x^2)
lnL = lim -(-1 - 4/x - 9/x^2)/(1 + 1/x + 2/x^2 + 3/x^3)
lnL = -(-1)/1 = 1
e^(lnL) = e^1
L = e