# Find this limit

1. May 24, 2013

### utkarshakash

1. The problem statement, all variables and given/known data

$\stackrel{lim}{x→0} \left( \dfrac{1^x+2^x+......n^x}{n} \right) ^{1/x}$

2. Relevant equations

3. The attempt at a solution
Let the quantity inside the bracket be represented by t.Rewriting

$(1+t-1)^{\frac{1}{t-1}.(t-1).\frac{1}{x}} \\ e^{(t-1)/x} \\ \stackrel{lim}{x→0} \left( \dfrac{1^x+2^x+......n^x-n}{nx} \right)$

Using L Hospital's Rule
$\stackrel{lim}{x→0} \left( \dfrac{x(ln1+ln2....ln n)}{n} \right)$

Now if I put x=0 I get limit as e^0 = 1. But this is not the correct answer.

2. May 24, 2013

### deluks917

if n= 1 our answer is as 1x = 1.

Take the ln of the expression. Then we get:

limx->0 ln((1x+...+nx)/n)/x

If we plug in x = 0 we get ln(1)/0 = 0/0 so we can use L'hopital's rule getting as d/dx x = 1:

limx->0 n/(1x+..._nx) * (ln(1)1n+...+ln(n)nx)/n =
limx->0 (ln(1)1x+...+ln(n)nx)/(1x+...+nx).

This is continuous at 0 so we can plug in x=0 to get:

(ln(1) + ... + ln(n))/(1+...+1) = (ln(1) + ... + ln(n))/(n).

So the original limit is exp((ln(1) + ln(2) ... + ln(n))/n)

edit: checked with matlab this is right.

Last edited: May 24, 2013
3. May 24, 2013

### Ray Vickson

You have a limit of the form
$$\lim_{x \to 0} \left( \frac{f_n(x)}{n}\right)^{1/x},\\ f_n(x) = \sum_{i=1}^n i^x$$
Never mind l'Hospital; just look at the form of $f_n(x)$ for small |x|, then use some familiar limit results.

4. May 24, 2013

### utkarshakash

I can't understand this step. Please tell me how you have differentiated the numerator?

5. May 25, 2013

### Pranav-Arora

What's the derivative of $a^x$ w.r.t x, where a is some constant?

6. May 25, 2013

### deluks917

I just canceled the n's. In latex:

$\frac{n}{1x+..._nx} \frac{ln(1)1n+...+ln(n)nx}{n}$

7. May 25, 2013

### deluks917

I just canceled the n's. In latex:

$\frac{n}{1^x+...+n^x} \frac{ln(1)1^n+...+ln(n)n^x}{n} = \frac{ln(1)1^n+...+ln(n)n^x}{1^x+...+n^x}$

8. May 25, 2013

### sankalpmittal

Approach 1: Take natural logarithm on both sides as, ln y = lim x->0 f(x).. Then you apply L' Hospital's Rule.

Approach 2: Write the given limit in the form of lim n→0 (1+n)1/n = e and then evaluate accordingly.

9. May 25, 2013

### utkarshakash

x lna

10. May 25, 2013

### dextercioby

No, it's ln a times a^x.

11. May 25, 2013

### Pranav-Arora

No.

dextercioby has already pointed out the correct derivative. Redo it.

12. May 25, 2013

### utkarshakash

Thanks for pointing out.

13. May 25, 2013

### utkarshakash

Can you type in the exact command that you used to find limit in matlab? It seems my command is giving me incorrect result.