when x near to infinity lim ((x+a)^(1+1/x)-x^(1+1/(x+a)))=?

Question

The answer is a ,and please use the hospital's rule to show, thanks.

Answer 1

(x+a)ⁿ=xⁿ+nax^n-1+... (binomial expansion)

(x+a)^1+1/x=x^1+1/x+(1+1/x)ax^1/x+...

When x→∞, 1/x→0, so (x+a)^1+1/x→x+a (as you might expect, because the exponent→1).

x^1+1/(x+a)→x as x→∞, so (x+a)^1+1/x-x^1+1/(x+a)→x+a-x=a.

l'Hôpital's Rule requires quotients of two functions f and g with their derivatives. We don't have the quotient of two functions.

Also, (x+a)^1+1/x=(x+a)(x+a)^1/x=x+a, x^1+1/(x+a)=x.x^1/(x+a)=x as x→∞. Hence the difference is a.