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Q: What is the limit of x divided by cot x?
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What is the exact trigonometric function value of cot 0?

There is no value cot 0, because cot 0 is equivalent to 1 / tan 0, which is equivalent to 1 / 0, which is undefined. That said, the limit of cot x as x approaches 0 is infinity.


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The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1


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The limit is 4.


How do you draw the graph of modulus of y equals cot x?

First note that this not the graph of y = |cot(x)|.The equivalent equations for |y| = cot(x) or cot(x) = |y| arecot(x) = -y or cot(x) = +ySo plot y = cot x and then reflect all the points in the x-axis.


What is the derivative of x equals cot2 divided by t?

Assuming you want dx/dt and that the equation is x = cot(2) / t (i.e. cot(2) is a constant) we can use the power rule. First, we rewrite it: cot(2)/t = cot(2) * t-1 thus, by the power rule: dx/dt = (-1) cot(2) * t-1 -1 = - cot(2) * t-2= = -cot(2)/t2


Second derivative of cosecx?

d/dx cosec(x) = - cosec(x) * cot(x) so the second derivative or d(d/dx)/dx cosec(x) = [- cosec(x) * d/dx cot(x)] + [ - d/dx cosec(x) * cot(x)] = [- cosec(x) * -cosec^2(x)] + [ - (- cosec(x) * cot(x)) * cot(x)] = cosec(x) * cosec^2(x) + cosec(x)*cot^2(x) = cosec(x) * [cosec^2(x) + cot^2(x)].


What is the infinite limit of 1 divided by ln x?

The limit should be 0.


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1


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You can use the L'hopital's rule to calculate the limit of e5x -1 divided by sin x as x approaches 0.


What is the derivative of cot x?

The derivative of cot(x) is -csc2(x).(Which is the same as -1/sin2(x).)


What is the second derivative of ln(tan(x))?

f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2


What is the integral of 1 divided by the sine squared of x with respect to x?

∫ 1/sin2(x) dx = -cot(x) + CC is the constant of integration.