lim x->pi cotx/cscx
To evaluate the limit of (cot(x) / csc(x)) as x approaches pi, we can simplify the expression by using trigonometric identities.
First, rewrite cot(x) and csc(x) in terms of sine and cosine:
cot(x) = cos(x) / sin(x)
csc(x) = 1 / sin(x)
Now, substitute these values back into the original expression:
(cot(x) / csc(x)) = (cos(x) / sin(x)) / (1 / sin(x))
Next, simplify further by multiplying by the reciprocal:
= (cos(x) / sin(x)) * (sin(x) / 1)
= cos(x)
Therefore, the limit of (cot(x) / csc(x)) as x approaches pi is equal to cos(pi).
Since cos(pi) = -1, the answer is -1.
To find the limit of (cotx/cscx) as x approaches pi, we can use the concept of L'Hopital's rule. L'Hopital's rule states that if we have a limit of the form f(x)/g(x), where both f(x) and g(x) approach 0 or infinity as x approaches the given value, then the limit of f(x)/g(x) as x approaches the given value is equal to the limit of f'(x)/g'(x) as x approaches the given value, provided this limit exists.
To apply L'Hopital's rule, we can rewrite the expression as (cosx/sinx)/(1/sinx). Now, as x approaches pi, sinx approaches 0 and cosx approaches -1. Thus, we have (-1/0)/(1/0), which is an indeterminate form of -∞/∞.
To find the limit, we can differentiate the numerator and denominator separately. The derivative of cosx is -sinx, and the derivative of sinx is cosx. So, the limit becomes (-(-sinx))/cosx = sinx/cosx = tanx.
Thus, the limit of (cotx/cscx) as x approaches pi is equal to tan(pi).
To find the limit of the function as x approaches pi:
lim x->pi (cotx/cscx)
We can simplify the expression using trigonometric identities. The cotangent function is the reciprocal of the tangent function, and the cosecant function is the reciprocal of the sine function. Using these identities:
cotx = 1/tanx
cscx = 1/sinx
Substituting these identities into the expression:
lim x->pi (cotx/cscx)
= lim x->pi ((1/tanx)/(1/sinx))
= lim x->pi (sinx/tanx)
Now, to evaluate this limit, we can use the fact that as x approaches pi, sinx approaches 0 and tanx approaches infinity. Therefore, the limit will be of the form (0/∞), which is an indeterminate form.
To resolve this indeterminate form, we can use L'Hôpital's Rule.
L'Hôpital's Rule states that if the limit of the ratio of two differentiable functions is indeterminate of the form (0/0) or (∞/∞), then we can take the derivative of the numerator and the denominator and re-evaluate the limit.
Differentiating the numerator and denominator:
d/dx (sinx) = cosx
d/dx (tanx) = sec^2x
Substituting these derivatives back into the expression:
lim x->pi (cosx/sec^2x)
Now, as x approaches pi, cosx approaches -1 and sec^2x approaches infinity. So, we have a new indeterminate form, (-1/∞).
Again, we can apply L'Hôpital's Rule:
d/dx (-1) = 0
d/dx (sec^2x) = 2sec^2x tanx
Substituting these derivatives back:
lim x->pi (0/(2sec^2x tanx))
= lim x->pi 0
Therefore, the limit of the given function as x approaches pi is 0.