Question

lim x to infinity[ln(1+x^2)-ln(1+x)]

Answer 1

L'hospitals Rule

Answer 2

\[\ln(1+x^2)-\ln(1+x)=\ln\left(\frac{1+x^2}{1+x}\right)\]

Answer 3

Or he could apply Limit Laws.

Answer 4

Or long division, for the fraction inside.

Answer 5

ln(1 + x^2) = ln(x^2(1 +1/x^2) = lnx^2 + ln(1 + 1/x^2)
ln(1 + x) = ln(x(1 + 1/x) = ln(x) + ln(1 + 1/x)
lim x-> infinity [ln(1 + x^2) - ln(1 + x)
lim x-> infinity [ln((x^2) - ln((x)
lim x-> infinity [ln((x) = infinity undefined