Help in limits maybe?

Here's a problem i am not being able to solve:

\(prove\) \(that\)

limx→π2(11cos2x+21cos2x+...........+n1cos2x)cos2x=n\large{lim_{x\to \frac{\pi}{2}}}\LARGE{(1^{\frac{1}{cos^{2}x}}+2^{\frac{1}{cos^{2}x}}+...........+n^{\frac{1}{cos^{2}x}})^{cos^{2}x}=n}

Please help

#Calculus #Limits #HelpMe! #Trigonometric

Note by Aritra Jana
6 years, 6 months ago

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Comments

To prove :

limx→π2(11cos2x+21cos2x+...........+n1cos2x)cos2x=n\displaystyle \large{lim_{x\to \frac{\pi}{2}}}\LARGE{(1^{\frac{1}{cos^{2}x}}+2^{\frac{1}{cos^{2}x}}+...........+n^{\frac{1}{cos^{2}x}})^{cos^{2}x}=n}

Let y=1cos2xy=\frac{1}{cos^2x} . As x→π2,y→∞x\to \frac{\pi}{2}, y\to \infty.

Therefore , limit=Limy→∞(1y+2y+3y+....+ny)1y∞0 formlimit=\displaystyle Lim_{y\to \infty} (1^y+2^y+3^y+....+n^y)^{\frac{1}{y}} \quad \quad \quad \infty^0 \ form =(ny)1y[(1n)y+(2n)y+(3n)y+......+(n−1n)y+1]1y\quad \quad \quad = (n^y)^{\frac{1}{y}} \left[ \left(\frac{1}{n} \right)^y+ \left(\frac{2}{n} \right)^y+ \left(\frac{3}{n} \right)^y+......+ \left(\frac{n-1}{n} \right)^y+1 \right]^{\frac{1}{y}} =n.[(1n)y+(2n)y+(3n)y+......+(n−1n)y+1]1y \quad \quad \quad = n.\left[ \left(\frac{1}{n} \right)^y+ \left(\frac{2}{n} \right)^y+ \left(\frac{3}{n} \right)^y+......+ \left(\frac{n-1}{n} \right)^y+1 \right]^{\frac{1}{y}} =n.(1)0=n\quad \quad \quad =n . (1)^0 =\boxed{n}

Enjoy @Aritra Jana !

Sandeep Bhardwaj - 6 years, 6 months ago

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ohhh.. silly me! i forgot to consider that each of in\frac{i}{n} for i≤n−1i≤n-1 is less than 11!!

making lim⁡y→∞(in)y=0\lim_{y\to\infty}(\frac{i}{n})^{y}=0

anyways. thanks a lot for replying :D

Aritra Jana - 6 years, 6 months ago
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