Question

evaluate lim x tends to 1 , [log(1-x^2)]/cotπx

Answer 1

you can see your answer

Answer 2

answer is 0

we have to evaluate $\displaystyle\lim_{x\to 1}\frac{log(1 - x^2)}{cot\pi x}$

putting, x = 1 + h

so, $\displaystyle\lim_{h+1\to 1}\frac{log(1 - (1+h)^2)}{cot\pi (1+h)}$

$\displaystyle\lim_{h\to 0}\frac{log(-h^2-2h)}{cot\pi h}$

$\displaystyle\lim_{h\to 0}\frac{tan\pi h}{log(-h^2-2h)}$

now above expression is in the form of 0/0. so we can apply L- Hospital rule.

$\displaystyle\lim_{h\to 0}\frac{\pi sec^2\pi h}{(-2h-2)/(-h^2-2h)}$

= $\displaystyle\lim_{h\to 0}\frac{(-h^2-2h)\pi sec^2\pi h}{2h-2}$

putting, h = 0

we get, {(0)² - 2(0))π sec²π(0)}/(2(0)-2) = 0

also read similar questions:lim x tends to 0 [(1+x)^1/x - e] / x is equal to

https://brainly.in/question/1225206

lim x tends to 0.

{(1^x+2^x+. . .+n^x)/n}^a/x

https://brainly.in/question/1508831