- #1

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[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 dont 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→∞