Feb 23, 2009 #2 F frank Member May 17, 2008 337 0 16 lim x/(x+1)^x (L'hôpital's rule) x-> oo = lim (x)'/((x+1)^x)' x-> oo n.b.: => y = (x+1)^x (used ln) ln y = x ln(x+1) y'/y = (x)' (ln(x+1)) + (ln(x+1))' (x) y'/y = ln(x+1) + (1/x+1)(x) y' = (x+1)^x [ln(x+1) + x(1/x+1)] =(**) lim 1/(x+1)^x [ln(x+1) + x(1/x+1)] x->oo n.b.: lim 1 +(1/x)^x = e x->oo so , lim 1/(**) = 1/e = e^-1 x-> oo
lim x/(x+1)^x (L'hôpital's rule) x-> oo = lim (x)'/((x+1)^x)' x-> oo n.b.: => y = (x+1)^x (used ln) ln y = x ln(x+1) y'/y = (x)' (ln(x+1)) + (ln(x+1))' (x) y'/y = ln(x+1) + (1/x+1)(x) y' = (x+1)^x [ln(x+1) + x(1/x+1)] =(**) lim 1/(x+1)^x [ln(x+1) + x(1/x+1)] x->oo n.b.: lim 1 +(1/x)^x = e x->oo so , lim 1/(**) = 1/e = e^-1 x-> oo
Feb 23, 2009 #3 F frank Member May 17, 2008 337 0 16 lim x/(x+1)^x (L'hôpital's rule) x-> oo = lim (x)'/((x+1)^x)' x-> oo n.b.: => y = (x+1)^x (used ln) ln y = x ln(x+1) y'/y = (x)' (ln(x+1)) + (ln(x+1))' (x) y'/y = ln(x+1) + (1/x+1)(x) y' = (x+1)^x [ln(x+1) + x(1/x+1)] =(**) lim 1/(x+1)^x [ln(x+1) + x(1/x+1)] x->oo n.b.: lim 1 +(1/x)^x = e x->oo so , lim 1/(**) = 1/e = e^-1 x-> oo
lim x/(x+1)^x (L'hôpital's rule) x-> oo = lim (x)'/((x+1)^x)' x-> oo n.b.: => y = (x+1)^x (used ln) ln y = x ln(x+1) y'/y = (x)' (ln(x+1)) + (ln(x+1))' (x) y'/y = ln(x+1) + (1/x+1)(x) y' = (x+1)^x [ln(x+1) + x(1/x+1)] =(**) lim 1/(x+1)^x [ln(x+1) + x(1/x+1)] x->oo n.b.: lim 1 +(1/x)^x = e x->oo so , lim 1/(**) = 1/e = e^-1 x-> oo