# Differentiation and the limit notation

I currently started studying some aspects of calculus for the first time in a very long time, and now I am struggling to understand some basic and silly concepts of calculus including the notations.

I have an example such as "Find $\lim_{n \to \infty} ( \log n^2 / n)$" -- which turns out to be $0$.

Now, is this question asking me to differentiate or I misunderstood it? If so, why is it using the limit notation rather than the differentiation notation: $d/dn$ or $f'(n)$. Also, are we always supposed to differentiate when we see something like "$\lim_{n \to \infty}$"?

Although this question might sound really silly, I would really appreciate some explanation.

Note: I forgot to say that-- it seems to me that the book did differentiate the numerator and the denominator in the last step which is why I got confused.

lim n→∞(log n^2/n) =lim n→∞(2log n/n) = lim n→∞((2/ln 2) (ln n /n) ) = (2/ln 2) lim n→∞(1/n) = 0.

The question has nothing to do with differentiation; it is asking for the limit as $n$ goes to infinity of $\frac{\log n^2} n$.

If you are not comfortable with limits, as it appears, I suggest you read about limits before reading about derivatives. With the new information in the question, your book is suggesting you apply L'Hospital's Rule. That's a bad idea if you are struggling with the basic concepts.

• But the book did differentiate the numerator and the denominator in the last step. lim n→∞(log n^2/n) =lim n→∞(2log n/n) = lim n→∞((2/ln 2) (ln n /n) ) = (2/ln 2) lim n→∞(1/n) = 0. It said "differentiate both numerator and the denominator" – Rain Feb 23 '17 at 6:14
• Please see the edit. – Martin Argerami Feb 23 '17 at 7:41
• Now, everything is clear. I actually know how to us the L'Hospital's Rule 'mechanically', but I was more confused with the usage of the limit and differentiation than I was with differentiation and limit rules. I really appreciate the help. – Rain Feb 23 '17 at 8:50

Consider infinity as a concept not a number. A diverging geometric progression -it gets bigger and bigger and never stops. Now, if you are to investigate an expression with a parameter with 'value' of infinity, you know to evaluate the expression as the parameter gets bigger and bigger because you cannot set a parameter equal to a never-ending progression.

Consider how to determine the rate of change of something. You differentiate with respect to some spatial or temporal component. If you want to determine how fast something is going, you differentiate the equation of displacement with respect to time. Here you want to see how fast the numerator grows, compared to the denominator. In other words, you want to compare the rates of change.

$$\lim_{n \to \infty} \frac{\log (n^2)}{n}$$

$$\frac{d}{dn} \frac{\log (n^2)}{n} = \frac{d}{dn}\left(\frac{2 log(n)}{n}\right)$$

Proceed to calculate using the quotient rule of differentiation:

$$\frac{n\frac{1}{n}-\log(n).1}{n^2} =\frac{1-log(n)}{n^2}$$

Nevertheless, logarithmic functions compress the value of a number, so log(n) will be less than n. This means that as n gets to be larger and larger, the denominator will make the fraction 'tend' towards zero. It may do so from the negative or positive 'side' of zero.

Check out l'hopital's rule for some interesting and informative mathematics.

You do not necessarily have to differentiate every time you see $\lim{n \to \infty}$, unless you are specifically interested in rate of change.

• This video on L'Hopital's rule should be interesting to you youtube.com/watch?v=P8o6QIODnWA – DWD Feb 22 '17 at 20:54
• This page on 'Radius of Convergence' should be interesting to you too tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx It basically measures whether as the parameter of a function becomes larger, whether the output of the expression diverges (get's larger) of converges (settles close to some number) – DWD Feb 22 '17 at 20:55
• Bro, I am really happy to see this explanation. You made my life much much easier. Thanks a lot :) – Rain Feb 23 '17 at 8:51
• My pleasure, once you've got this down, you'll be unstoppable! Unless you converge as t tends toward infinity that is ;) – DWD Feb 23 '17 at 10:54