Find Limit: Homework Statement & Equations

[utkarshakash]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Let the quantity inside the bracket be represented by t.Rewriting

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

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.

 

[deluks917]

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

Take the ln of the expression. Then we get:

lim_x->0 ln((1^x+...+n^x)/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:

lim_x->0 n/(1^x+..._n^x) * (ln(1)1ⁿ+...+ln(n)n^x)/n =
lim_x->0 (ln(1)1^x+...+ln(n)n^x)/(1^x+...+n^x).

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.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Let the quantity inside the bracket be represented by t.Rewriting

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

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.

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

 

[utkarshakash]

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


lim_x->0 n/(1^x+..._n^x) * (ln(1)1ⁿ+...+ln(n)n^x)/n =
lim_x->0 (ln(1)1^x+...+ln(n)n^x)/(1^x+...+n^x).


edit: checked with MATLAB this is right.

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

 

[Saitama]

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

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

 

[deluks917]

I just canceled the n's. In latex:

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

 

[deluks917]

I just canceled the n's. In latex:

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

 

[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.

 

[utkarshakash]

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

x lna

 

[dextercioby]

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

 

[Saitama]

x lna

No.

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

 

[utkarshakash]

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

Thanks for pointing out.

 

[utkarshakash]

edit: checked with MATLAB this is right.

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