Evaluate the following

Question

Lim _{x  →∞}  [ x / 1 + x ] ^x

Evaluate

Answer 1

when you have a limit where the base and exponent both have a variable, we need to use a log, then L'hospitals Rule

so first let's do ln of our limit to get x out of the exponent

ln [lim x->oo (x/1+x)^x] = x * ln[lim x->oo x/1+x] = lim x->oo[x*ln(x/1+x)]

then we can rewrite it so we get a fraction

lim x->oo[ln(x/1+x) / (1/x)]

Now if we do limit, the limit as x goes to infinity of x/1+x is 1 and ln1 is 0, and limit of 1/x also goes to 0, so it's 0/0, and we can use L'hospitals Rule (which means we do derivative of top and bottom)

derivative of ln(x/1+x) = 1/(x/1+x) * [(1+x)*(1)-(x)*(1) / (1+x)^2] = (1+x)/x * 1/(1+x)^2 = (1+x)/x(x+1)^2 = 1/x(x+1)

derivative of 1/x is -1/x^2

so now our limit is:

lim x->oo [1/x(x+1)] / [-1/x^2] = lim x->oo (-x^2)/x(x+1) = lim x->oo (-x^2)/(x^2+x) = -1

but then remember we did ln of this whole limit, so we have to undo that by raising e to our answer

so our end limit = e^-1 = 1/e