limits of ln?

Question

CALCULUS: LIMITS o' Natural Log!!?
lim as x approaches infinity (+) ...ln [(x^.5) + 5] divided by ln x...i believe it is 1/2...i may be wrong please help...i need by tomorrow !! no lopital plz

Answer 1

If you are solving the limit for ln [(x^(1/2)) + 5]/lnx, you're right. Just reason as follows:

x^(1/2) is increasing to infinity on the interval (0,inf). Hence, eventually it is bigger than 5. Since ln(x) is also increasing, and (x^(1/2)) + 5 < (x^(1/2)) + (x^(1/2)) = 2(x^(1/2)). Hence
ln [(x^(1/2)) + 5]/lnx < ln [2(x^(1/2))]/lnx = (ln2 +(1/2)lnx)/lnx > ln2/lnx + (1/2)-->1/2 ***by the properties of ln***.

For the other direction, note that eventually 5>-(1/2)x^(1/2). Hence ln [(x^(1/2)) + 5]/lnx > ln [(x^(1/2)) -(1/2)x^(1/2]/lnx = (-ln2 +(1/2)lnx)/lnx-->1/2 as x-->inf. It follows by the pinching theorem that ln [(x^(1/2)) + 5]/lnx -->1/2, just as you predicted!


Just in case the question had power 5 instead of 0.5:
Since lnx is increasing on (0, inf) for all x>5^(1/5), x^5>5 and therefore ln(x^5+5)
ln[(x^5)+5]/lnx < ln(2x^5)/ln(x)= (ln2+5ln(x))/ln(x)-->5

On the other hand, for all x>10^(1/5), 5 > -(1/2)x^5. Hence ln[(x^5)+5]/lnx > ln[(x^5)-(1/2)x^5]/lnx = ln[(1/2)(x^5)]/lnx = (-ln2 +5lnx)/lnx-->5. It follows by the pinching theorem that ln[(x^5)+5]/lnx-->5

Answer 2

there is not any finite decrease. as n is going to infinity, so does log n, albeit slowly. elevating to the two hundredth skill would not change this . We might want to coach that for any ok > 0, there exists N such that n > N implies (ln(n))^2 hundred > ok on condition that ln(n) is monotone increasing, it suffices to exhibit that f(N) > ok. nicely, given ok, go with N = max(e^2, e^(ok+a million)) f(N) = (ln(N))^2 hundred = (ln(e^(ok+a million)))^2 hundred = (ok+a million)^2 hundred > ok+a million > ok e^2s in simple terms in there in case ok

Answer 3

infinite/infinite

indefinite

use L'Hopital rule

Take derivative of numerator and denominator.