# Trigonometric limit?

• Jun 9th 2009, 08:20 AM
fardeen_gen
Trigonometric limit?
Evaluate:

$\lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x}$
• Jun 9th 2009, 11:14 AM
sinewave85
Quote:

Originally Posted by fardeen_gen
Evaluate:

$\lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x}$

It would seem that:

$\lim_{x\rightarrow 0+} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = (+\infty)^{0} = 1$

and

$\lim_{x\rightarrow 0-} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = (-\infty)^{0} = 1$

so

$\lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = 1$
• Jun 9th 2009, 11:32 AM
Moo
Quote:

Originally Posted by sinewave85
It would seem that:

$\lim_{x\rightarrow 0+} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = (+\infty)^{0} = 1$

But this is not defined (Worried)

Quote:

and

$\lim_{x\rightarrow 0-} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = (-\infty)^{0} = 1$
How can something positive have a negative limit ?

Quote:

so

$\lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = 1$
...
• Jun 9th 2009, 11:36 AM
sinewave85
Quote:

Originally Posted by Moo
But this is not defined (Worried)

How can something positive have a negative limit ?

...

Very sorry -- I forgot that $\infty^{0}$ was undefined. Thanks for catching that, Moo. I would hate to lead someone astray.
• Jun 9th 2009, 09:34 PM
fardeen_gen
How do we deal with $\infty^{0}$ forms? This problem has been giving me a headache.
• Jun 10th 2009, 03:55 AM
Ruun
$\frac{\sin^2(x)}{\sin^2(x)} (\frac{1}{\sin^2(x)}+\frac{1}{2\sin^2(x)}+ \mbox(...) +\frac{1}{n \sin^2(x)})^{\sin^2(x)}=\frac{1}{\sin^2(x)}(H_n)^{ \sin^2(x)}=L$

$ln (L) = ln(H_n)$

Harmonic number - Wikipedia, the free encyclopedia
• Jun 10th 2009, 04:32 AM
Isomorphism
Quote:

Originally Posted by fardeen_gen
Evaluate:

$\lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x}$

Using the fact that $\sin x \approx x$, for small x, we can write:

$\lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} =$
$\lim_{t \rightarrow \infty} \left(1^{t^2} + 2^{t^2} + 3^{t^2} + \mbox{...} + n^{t^2}\right)^{\frac1{t^2}}$

Taking logs and apply LHospital's rule, we have to evaluate $\lim_{t \rightarrow \infty} \frac{\log \left(1^{t^2} + 2^{t^2} + 3^{t^2} + \mbox{...} + n^{t^2}\right)}{t^2}$ to get $\log n!$, Finally the answer is n!
• Jun 11th 2009, 11:40 AM
Random Variable
Quote:

Originally Posted by Isomorphism
Using the fact that $\sin x \approx x$, for small x, we can write:

$\lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} =$ $\lim_{t \rightarrow \infty} \left(1^{t^2} + 2^{t^2} + 3^{t^2} + \mbox{...} + n^{t^2}\right)^{\frac1{t^2}}$

Taking logs and apply LHospital's rule, we have to evaluate $\lim_{t \rightarrow \infty} \frac{\log \left(1^{t^2} + 2^{t^2} + 3^{t^2} + \mbox{...} + n^{t^2}\right)}{t^2}$ to get $\log n!$, Finally the answer is n!

After one application of L'Hospital's rule, I get

$\lim_{t \to \infty} \frac {\ln(1)1^{t^{2}} + \ln(2)2^{t^{2}} + \ln(3)3^{t^{2}} + ... + \ln(n)n^{t^{2}}}{1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}}}$

then it would appear you did the following

$\lim_{t \to \infty} \frac {\big(\ln(1) +\ln(2) + \ln(3) + ... + \ln(n)\big)(1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}})}{ 1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}}}$ ?

$= \lim_{t \to \infty} \ln(1*2*3*...*n)$ which would be correct if the previous step were correct

What am I missing?
• Jun 11th 2009, 09:38 PM
Isomorphism
Quote:

Originally Posted by Random Variable
After one application of L'Hospital's rule, I get

$\lim_{t \to \infty} \frac {\ln(1)1^{t^{2}} + \ln(2)2^{t^{2}} + \ln(3)3^{t^{2}} + ... + \ln(n)n^{t^{2}}}{1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}}}$

then it would appear you did the following

$\lim_{t \to \infty} \frac {\big(\ln(1) +\ln(2) + \ln(3) + ... + \ln(n)\big)(1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}})}{ 1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}}}$ ?

$= \lim_{t \to \infty} \ln(1*2*3*...*n)$ which would be correct if the previous step were correct

What am I missing?

(Tmi) You are not missing anything...Its wrong and was hastily written(I applied $t \to 0$)

Here is the solution:

$\lim_{t \to \infty} \frac {\ln(1)1^{t^{2}} + \ln(2)2^{t^{2}} + \ln(3)3^{t^{2}} + ... + \ln(n)n^{t^{2}}}{1^{t^{2}} + 2^{t^{2}} + 3^{t^{2}} + ... + n^{t^{2}}}$

$= \lim_{t \to \infty} \dfrac {0 + \ln(2)\left(\frac2{n}\right)^{t^{2}} + \ln(3)\left(\frac3{n}\right)^{t^{2}} + ... + \ln(n)}{\left(\frac1{n}\right)^{t^{2}} + \left(\frac2{n}\right)^{t^{2}} + \left(\frac3{n}\right)^{t^{2}} + ... + 1}$

$=\ln(n)$