`lim_(x->(pi/2)) cos(x)/(1 - sin(x))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t...

`lim_(x->(pi/2)) cos(x)/(1 - sin(x))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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Textbook Question

Chapter 4, 4.4 - Problem 11 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the limit, hence, you need to replace `pi/2 ` for x such that:

`lim_(x->pi/2) (cos x)/(1 - sin x) = (cos (pi/2))/(1 - sin (pi/2)) = 0/(1-1) = 0/0`

Since the limit is indeterminate `0/0,` you may apply l'Hospital's rule:

`lim_(x->pi/2) (cos x)/(1 - sin x) = lim_(x->pi/2) ((cos x)')/((1 - sin x)') `

`lim_(x->pi/2) ((cos x)')/((1 - sin x)')= lim_(x->pi/2) (-sin x)/(-cos x)`

Replacing by `pi/2 ` yields:

`lim_(x->pi/2) (-sin x)/(-cos x) = lim_(x->pi/2) tan x = tan (pi/2)`

If `x->pi/2 ` and `x<pi/2` , then `tan (pi/2) = +oo`

If `x->pi/2 ` and `x>pi/2` , then `tan (pi/2) = -oo`

Hence, evaluating the limit, yields` lim_(x->pi/2) (cos x)/(1 - sin x) = +oo` o if` x->pi/2` and `x<pi/2` , or `lim_(x->pi/2) (cos x)/(1 - sin x) = -oo` , if `x->pi/2` and `x>pi/2.`

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