2 Answers

Rami Ma · Answer 1 · 2020-07-09T23:09:04+0000

Not sure why my previous answer was deleted...

If you substitute
$y=\frac1{\sin^2x}=\csc^2x$ , then
$x\to0$ gives
$y\to+\infty$ , so the limit is equal in value to

$\displaystyle\lim_(y\to\infty)\left(1^y+2^y+\cdots+n^y\right)^(1/y)$

Since
$e^(\ln x)=x$ , we can write

$\left(1^y+2^y+\cdots+n^y\right)^(1/y)=e^{\ln(1^y+2^y+\cdots+n^y)^(1/y)=e^(\ln(1^y+2^y+\cdots+n^y)/y)$

Because
e^x is continuous at all
, we can pass the limit to the argument of the exponential function. That is,

$\displaystyle\lim_(x\to\infty)e^(f(x))=e^{\lim\limits_(x\to\infty)f(x)}$

so that the limit we're interested in is equal to

$e^{\lim\limits_(y\to\infty)\ln(1^y+2^y+\cdots+n^y)/y}$

For all natural numbers
, we have

$1^y+2^y+\cdots+n^y\le n^y+n^y+\cdots+n^y=n\cdot n^y=n^(y+1)$

so

$\displaystyle\lim_(y\to\infty)\frac{\ln(1^y+2^y+\cdots+n^y)}y=\lim_(y\to\infty)\frac{\ln n^(y+1)}y=\ln n\cdot\lim_(y\to\infty)\frac{y+1}y=\ln n$

and this makes the overall limit take on a value of

$e^(\ln n)=\boxed n$

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