# Need working checked for limits question

1. Mar 10, 2005

### shan

My algebra's shocking so I'm not sure if I found the limit correctly... (sorry, I can't figure out this latex thing :grumpy: )

lim n tends to infinity of (sqrt(n))^-n

Using the rules of indices, I simplified it so that it became
n^(-n/2)

And then I used the log rule thing on it.

e^ lim (-n/2 * ln(n))

This is where I'm not sure: lim (-n/2 * ln(n))
I took the limit of the numerator (-n * ln(n)) to be infinity as n tends to infinity and the limit of the denominator would be 2. Infinity/2 is infinity so e^infinity is infinity...

Could someone please confirm or correct? Thanks :)

2. Mar 10, 2005

### xanthym

LIM{n → ∞} (√n)(-n) = LIM{n → ∞} 1/{(√n)(n)} =
= LIM{n → ∞} 1/{(n)(n/2)} =
-----> 1/∞ -----> 0
LIM{n → ∞} (√n)(-n) = 0

~~

3. Mar 10, 2005

### cepheid

Staff Emeritus
As xanthym showed you, it was not necessary to rewrite the expression as a power of e in order to evaluate the limit.

That having been said, there's nothing wrong with doing so per se, it's just an unnecessary extra manipulation. You had it almost all right. Your only error was here:

Wouldn't that be negative infinity? And so:

$$e^{-\infty} = \frac{1}{e^{\infty}} = 0$$

I just thought I would point that out, since I always find it helpful to see not only the correct solution, but where exactly I went wrong in my attempt. You should have been suspicious when you got the result that: $\frac{1}{(\sqrt{n})^n}$ increases as n increases. It's clear the opposite must be true.

4. Mar 11, 2005

### shan

Thanks very much cepheid for explaining about the negative infinity thing :) We're supposed to use these 'rules' we were taught to solve the problem, one of which is finding the limit of the logarithm first.

Anyway... sorry to be a pain but there's another question I'm not too sure on ^^;

lim (n tends to infinity) of n^3/5^n

As it is indeterminant, I used l'hopital's rule and differentiated the top and bottom.
(3n^2)/(n*5^(n-1))
I'm not too sure if I can differentiate 5^n to n*5^(n-1) using the indice method though...
but I simplified the above to
3n/5^(n-1)
and because it is an indeterminant, I used l'hopital's rule again.
3/((n-1)*5^(n-2)
The limit of that as n tends to infinity would be 0.

My big question is... Can I differentiate 5^n to n*5^(n-1) ?

5. Mar 11, 2005

### cepheid

Staff Emeritus
Nooooo.... :yuck:

Okay: what kind of function is this?

f(x) = xc

c = const.

It is a power function, right? Because the variable (x) is the base, and it is being raised to some constant exponent c. Being a power function, it can be differentiated using the power rule:
d/dx (xc) = cxc-1
______________________________________________________________
Now, what kind of function is this?

f(x) = cx

c = const.

Well, that's wacky! Now the base is a constant, and the variable (x) is in the exponent! This is not a power function at all! It is a totally different type of function. Not surprisingly, it is an exponential function. Do you remember how to differentiate exponential functions such as f(x) = 5x (the one in your problem)?

Hint: the answer has something to do with our favourite exponential function: f(x) = ex

You must be sure to distinguish between these two types of functions, as they are completely unrelated. So...just to be sure:

f(x) = x2 is a(n) ___________ function

g(x) = 3x is a(n) ___________ function

Last edited: Mar 11, 2005
6. Mar 11, 2005

### shan

lol ok point taken :)

So x^c is a power function and c^x is an exponential function...

So if I differentiated 5^n, it would still be 5^n?

7. Mar 11, 2005

### cepheid

Staff Emeritus
No...what's the definition of a derivative? How would you apply it here?

I'd recommend looking this up. It should be covered in your calculus book.

8. Mar 11, 2005

### dextercioby

For simplicity
$$(5^{x})'=[(e^{\ln 5})^{x}]'=...$$

Daniel.

9. Mar 11, 2005

### shan

Thanks guys, I found it on the net afterall ^^