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LenaWriter [7]
3 years ago
8

Lim x→π/2 1-sinx/cot^2xany genious help please ​

Mathematics
1 answer:
Simora [160]3 years ago
4 0

Rewrite the limand as

(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = (1 - sin(<em>x</em>)) / (cos²(<em>x</em>) / sin²(<em>x</em>))

… = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / cos²(<em>x</em>)

Recall the Pythagorean identity,

sin²(<em>x</em>) + cos²(<em>x</em>) = 1

Then

(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / (1 - sin²(<em>x</em>))

Factorize the denominator; it's a difference of squares, so

1 - sin²(<em>x</em>) = (1 - sin(<em>x</em>)) (1 + sin(<em>x</em>))

Cancel the common factor of 1 - sin(<em>x</em>) in the numerator and denominator:

(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = sin²(<em>x</em>) / (1 + sin(<em>x</em>))

Now the limand is continuous at <em>x</em> = <em>π</em>/2, so

\displaystyle\lim_{x\to\frac\pi2}\frac{1-\sin(x)}{\cot^2(x)}=\lim_{x\to\frac\pi2}\frac{\sin^2(x)}{1+\sin(x)}=\frac{\sin^2\left(\frac\pi2\right)}{1+\sin\left(\frac\pi2\right)}=\boxed{\frac12}

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