what is the limit of lim x->p/2 (tan(x-(p/2)))/(x-(p/2)+cos(x))note that x approach to (p/2) and lim x->p/2 (tan(x-(p/2)))/(x-(p/2)+cos(x)) not lim x->p/2 (tan(x-(p/2)))/(x-(p/2)) + cos(x)



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thilina-g's profile pic

Posted on (Answer #1)


If you try to evaluate the limit straight away, you get an indeterminate answer 0/0. In such situations you can use the l'hopitals rule.

`lim_(x-gta)(f(x))/(g(x)) = lim_(x-gta)(f'(x))/(g'(x))`

`f(x) = tan(x-pi/2)`

So, `f'(x) = sec^2(x-pi/2)`

`g(x) = (x-pi/2)+cos(x)`

`g'(x) = 1-sin(x)`



`lim_(x-gtpi/2)tan(x-pi/2)/((x-pi/2)+cos(x)) = lim_(x->pi/2)(sec^2(x-pi/2))/(1-cos(x))`

`lim_(x->pi/2)(sec^2(x-pi/2))/(1-cos(x)) = 1/(1-1) = +oo`


`lim_(x-gtpi/2)tan(x-pi/2)/((x-pi/2)+cos(x)) = +oo`


pourjour's profile pic

Posted on (Answer #2)

what about the right 0 and left of 0. How did you know the sign of 0 to say +infini or -infini ?
thilina-g's profile pic

Posted on (Answer #3)

There is a small mistake there.

It should be corrected as follows,

`lim_(x-gtpi/2)tan(x-pi/2)/((x-pi/2)+cos(x)) = lim_(x->pi/2)(sec^2(x-pi/2))/(1-sin(x))`

` lim_(x->pi/2)(sec^2(x-pi/2))/(1-sin(x)) = 1/(1-1) = +oo`


`lim_(x-gtpi/2)tan(x-pi/2)/((x-pi/2)+cos(x)) = +oo`



Now if you look at (1-sin(x)), the maximum sin(x) can achieve is 1, that is at pi/2, but when x is approaching to pi/2, sin(x) is not 1, it is a value little below 1. So (1- sin(x)) is positive. Therefore the limit is + infinity.



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