$\lim_{x\to 0}\left[1^{\frac{1}{\sin^2 x}}+2^{\frac{1}{\sin^2 x}}+3^{\frac{1}{\sin^2 x}}+\cdots + n^{\frac{1}{\sin^2 x}}\right]^{\sin^2x}$

Question

$$\lim_{x\to 0}\left[1^{\frac{1}{\sin^2x}}+2^{\frac{1}{\sin^2x}}+3^{\frac{1}{\sin^2x}}+\cdots + n^{\frac{1}{\sin^2x}}\right]^{\sin^2x}$$

Limit is of form $(\infty)^{0} $ $$\lim_{x\to 0}e^{\sin^2x\log{ {\left[1^{\frac{1}{\sin^2x}}+2^{\frac{1}{\sin^2x}}+3^{\frac{1}{\sin^2x}}+\cdots + n^{\frac{1}{\sin^2x}}\right]}}}$$

I don't know how to proceed further.

Easier to see that if $x\to 0$ then $y=1/\sin^2 x\to +\infty$, and try to compute: $$\lim_{y\to +\infty} \left[1^y+2^y+\cdots+n^y\right]^{1/y}$$ The $\sin^2 x$ part is essentially an irrelevant complication. — Thomas Andrews, Oct 01 '16 at 16:02
@ThomasAndrews Yeah! Question was really easy .Thanks for help. — Aakash Kumar, Oct 01 '16 at 16:05

score 4 · Accepted Answer · answered Oct 01 '16 at 16:06

4

Hint:

$$n^y\leq 1^y+2^y+3^y+\cdots + n^y\leq n^y+n^y+\cdots+n^y=n^{y+1}$$

answered Oct 01 '16 at 16:06

Thomas Andrews

177,126

I think taking $n^y $ common from all term will be enough. – Aakash Kumar Oct 01 '16 at 16:08
I don't understand why you have written this inequality. – Aakash Kumar Oct 01 '16 at 16:10
I find using the squeeze lemma simpler here, but to each their own. @AakashKumar – Thomas Andrews Oct 01 '16 at 16:10
If $f(x)\leq g(x)\leq h(x)$ and $f(x)\to L$ and $h(x)\to L$ then by the squeeze lemma, $g(x)\to L$. So raise each term here to $1/y$ then note that $(n^y)^{1/y}=n\to n$ and $(n^{y+1})^{1/y}=n^{1+1/y}\to n$. @AakashKumar – Thomas Andrews Oct 01 '16 at 16:12
This is a very nice way forward! Good to see you're back Thomas. -Mark – Mark Viola Oct 01 '16 at 16:50
Never left, @Dr.MV – Thomas Andrews Oct 01 '16 at 16:53

Mark Viola · Answer 2 · 2016-10-01T16:45:23.453

Note that we can write

$$\begin{align} \lim_{x\to 0}\left(\sum_{k=1}^n k^{\csc^2(x)}\right)^{\sin^2(x)}&=n\lim_{x\to 0}\left(\sum_{k=1}^n (k/n)^{\csc^2(x)}\right)^{\sin^2(x)} \tag 1\\\\ &=n \tag 2 \end{align}$$

In going from $(1)$ to $(2)$ we proceeded by evaluating the limit

$$\begin{align} \lim_{x\to 0}\left(\sum_{k=1}^n (k/n)^{\csc^2(x)}\right)^{\sin^2(x)} &=\lim_{x\to 0}e^{\sin^2(x)\log\left(1+\sum_{k=1}^{n-1}(k/n)^{\csc^2(x)}\right)}\\\\ &=\lim_{x\to 0}e^{\sin^2(x)\,O\left(\sum_{k=1}^{n-1}(k/n)^{\csc^2(x)}\right)}\\\\ &=e^0\\\\ &=1 \tag 2 \end{align}$$

score 3 · Answer 3 · answered Oct 01 '16 at 16:56

3

$$ \begin{align} &\lim_{x\to0}\left[1^{\frac1{\sin^2(x)}}+2^{\frac1{\sin^2(x)}}+3^{\frac1{\sin^2 (x)}}+\cdots+n^{\frac1{\sin^2(x)}}\right]^{\sin^2(x)}\\ &=\lim_{x\to\infty}\left[1^x+2^x+3^x+\cdots+n^x\right]^{1/x}\\ &=n\lim_{x\to\infty}\left[\left(\frac1n\right)^x+\left(\frac2n\right)^x+\left(\frac3n\right)^x+\cdots+\left(\frac{n-1}n\right)^x+1^x\right]^{1/x}\\[4pt] &=n\,[0+0+0+\cdots+0+1]^0\\[8pt] &=n \end{align} $$

answered Oct 01 '16 at 16:56

robjohn

345,667

A bit like mine. ;-)) – Mark Viola Oct 01 '16 at 16:58

score 1 · Answer 4 · answered Oct 01 '16 at 16:14

1

This is not in the form of$(\infty)^0$ as $1^{\infty}$ is undetermined, so try this process.

answered Oct 01 '16 at 16:14

sai saandeep

1,145

robjohn sir, you have done similar to my process – sai saandeep Oct 04 '16 at 16:07

$\lim_{x\to 0}\left[1^{\frac{1}{\sin^2 x}}+2^{\frac{1}{\sin^2 x}}+3^{\frac{1}{\sin^2 x}}+\cdots + n^{\frac{1}{\sin^2 x}}\right]^{\sin^2x}$

4 Answers4