- #1
damoj
- 9
- 0
basically i have to check if
[tex] xln\frac{(x+1)}{x} [/tex]→ 1 as x→∞
the first term is 0 as x→∞
in the answers they say they used maclaurin series and got
[tex]x(\frac{1}{x} + O\frac{1}{1^{2}}) [/tex]
but don't show how they did it.
would the first term in the series be
[tex]a(ln(\frac{a+1}{a}))[/tex]
there a = 0, but then the ln function is defined?
im getting a little confused about have a=0 but we want to figure out where the function goes to as x→∞
[tex] xln\frac{(x+1)}{x} [/tex]→ 1 as x→∞
the first term is 0 as x→∞
in the answers they say they used maclaurin series and got
[tex]x(\frac{1}{x} + O\frac{1}{1^{2}}) [/tex]
but don't show how they did it.
would the first term in the series be
[tex]a(ln(\frac{a+1}{a}))[/tex]
there a = 0, but then the ln function is defined?
im getting a little confused about have a=0 but we want to figure out where the function goes to as x→∞