hard limit problem?

Question

could someone give me at least the steps or how to get the steps to doing this problem?

lim as x goes to infinity of the function

(ln((square root of x) +5)) / ln(x)

Note it is not the square root of the quantity (x+5); it is only the square root of x then the plus five is outside of the square root.

Answer 1

sqrt(x) + 5 = sqrt(x) [1 + 5/sqrt(x)]

ln(sqrt(x) + 5) = ln(sqrt(x) [1 + 5/sqrt(x)]) = ln(sqrt(x)) + ln(1 + 5/sqrt(x))

Answer 2

Applying L'Hopital's Rule, you get

lim x-->inf ((1/(sqrt(x) +5) *1/(2sqrt(x))) / (1/x)

Simplifying:

lim x-->inf x/(2x + 10sqrt(x))

Applying LR again,

lim x-->inf 1/(2 + 5/sqrt(x)) = 1/2

Answer 3

because all of us understand a^x = a million +x.log(a) +(x^2/2!)*(log a)^2+...........................as much as infinity subtract a million from the two ingredient (a^x) -a million = x.log(a) ++(x^2/2!)*(log a)^2+...........................as much as infinity Divide the two ingredient by utilizing x { (a^x) -a million}/x = log(a) ++(x/2!)*(log a)^2+ greater powers of x...........................as much as infinity Taking shrink x->0, we get { (a^x) -a million}/x = log(a) +0+0+0+...........................as much as infinity { (a^x) -a million}/x = log(a) right here a = 2, consequently { (2^x) -a million}/x = log(2) .....................................Ans