Question

limt ln (n+1)/ln n=?
n->infty
evaluate without using L. hopital rule.

Answer 1

(1/(n+1))/(1/n)
1/(n+1)*n/1
n/(n+1)

Answer 2

^ Without....

Answer 3

As n appraoches infinity then its equal to 0

Answer 4

ya i just don't know how to write out the limit sign

Answer 5

It's equal to one...

Answer 6

ya u r right

Answer 7

i just guessed that one

Answer 8

I guess rosy's gone

Answer 9

why wouldn't you use L'Hopitals rule here ??
yes the limit is 1

Answer 10

There are other ways to do it. Like the following:

Answer 11

cus she most probably didnt learn it yet

Answer 12

\[\lim_{n \rightarrow \infty}\frac{\ln(n+1)}{\ln(n)} \]
let
\[n = e^x\]
for some x. Then,
\[n+1 = e^x + 1 = e^x(1+e^{-x}) \]

making those substitutions, our expression becomes
\[\frac{\ln(e^x(1+e^{-x}))}{\ln(e^x}= \frac{\ln(e^x) + \ln(1+e^{-x})}{\ln(e^x)} \]
this simplifies to
\[1 + \frac{\ln(1+e^{-x})}{x} \]
Notice that as n approaches infinity, x approaches infinity -- so we can restate our problem as
\[ \lim_{x \rightarrow \infty} \quad 1 + \frac{\ln(1+e^{-x})}{x} = 1 + \frac{0}{\infty} = 1\]
No calculus required. (Forgive the abuse of notation with the zero over infinity bit...)

Answer 13

ya that was abetter explanation thn i give. that is the explanation that she must need

Answer 14

I suspect she probably has learned l'Hospital's rule, which is why she was specifically directed not to use it. It's a fast and easy method to compute limits, but it's not the only way, and algebraically evaluating complicated limits is an excellent exercise in thinking across the boundaries of mathematical disciplines. I know calculus cold but it even took me a few minutes of thinking to come up with a tricky way to do it, and I think that's the point.

Answer 15

oh ya what level of math r u in?

Answer 16

I'm graduating with a double major in physics and classical mathematics in about two weeks.

Answer 17

oh so ya i dont think i shld mess with u!!!

Answer 18

:) There are many more knowledgeable mathematicians than me hanging around this website.

Answer 19

ya but once they are too knowledgable they no longer have time to help u

Answer 20

0/infinity is an undetermined form isn't it? then how can u solve it by this method...

Answer 21

Jemurray 3 i think your method of solving my que is wrong how can you say o/infinity is zero? explain this if u r right

Answer 22

No, it's not indeterminate. Infinity over infinity or zero over zero is indeterminate, but zero over infinity is zero.

Answer 23

All of this is horrifically non-rigorous and heuristic, but I can be more math-y if you'd prefer.

Answer 24

Example.
\[\lim_{x \rightarrow \infty} \frac{1}{x} = 0\]
do you agree? In that case, the numerator is staying the same and the denominator is getting very large. Therefore, the fraction decreases to zero.
In your case, not only is the denominator getting very large, but the numerator is getting very small. If you would prefer, you can write it as
\[ \lim_{x\rightarrow \infty} \space \ln(1+e^{-x}) \cdot \frac{1}{x}\]
both terms converge to zero, so of course their product converges to zero.

Answer 25

in short you can say 0/infinity=o/1/0=0/1=0

Answer 26

My conscience demands that I stress that infinity is not a real number and anything divided by zero is not defined and is strictly speaking nonsensical. However, speaking in terms of limits,
as the denominator of a fraction grows and grows and grows while the numerator stays the same, the whole fraction will get closer and closer to zero.

Answer 27

So you need to be very careful in how you think about infinity. I stress that it's NOT a number, and we say something approaches infinity when it grows larger and larger without bound. For instance, things like
\[ \infty \cdot \frac{1}{\infty} = 1 \]
or
\[0\cdot \frac{1}{0} = 1 \]
make absolutely no sense mathematically so you cannot think of them in this way.