1. ## Trigonometric limit?

Evaluate:

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

2. Originally Posted by fardeen_gen
Evaluate:

$\displaystyle \lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x}$
It would seem that:

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

and

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

so

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

3. Originally Posted by sinewave85
It would seem that:

$\displaystyle \lim_{x\rightarrow 0+} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x} = (+\infty)^{0} = 1$
But this is not defined

and

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

so

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

4. Originally Posted by Moo
But this is not defined

How can something positive have a negative limit ?

...
Very sorry -- I forgot that $\displaystyle \infty^{0}$ was undefined. Thanks for catching that, Moo. I would hate to lead someone astray.

5. How do we deal with $\displaystyle \infty^{0}$ forms? This problem has been giving me a headache.

6. $\displaystyle \frac{\sin^2(x)}{\sin^2(x)} (\frac{1}{\sin^2(x)}+\frac{1}{2\sin^2(x)}+ \mbox(...) +\frac{1}{n \sin^2(x)})^{\sin^2(x)}=\frac{1}{\sin^2(x)}(H_n)^{ \sin^2(x)}=L$

$\displaystyle ln (L) = ln(H_n)$

Harmonic number - Wikipedia, the free encyclopedia

7. Originally Posted by fardeen_gen
Evaluate:

$\displaystyle \lim_{x\rightarrow 0} \left(1^{\frac{1}{\sin^2 x}} + 2^{\frac{1}{\sin^2 x}} + 3^{\frac{1}{\sin^2 x}} + \mbox{...} + n^{\frac{1}{\sin^2 x}}\right)^{\sin^2 x}$
Using the fact that $\displaystyle \sin x \approx x$, for small x, we can write:

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

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

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

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

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

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

then it would appear you did the following

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

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

What am I missing?

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

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

then it would appear you did the following

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

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

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

Here is the solution:

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

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

$\displaystyle =\ln(n)$