limits question?

Question

the limit as x approaches pi/2 of (x-pi/2)tanx

I don't do very well with radians, and i'm not exactly sure how to work this one.......i thought of maybe changing tanx to (sin/cos)x and distributing? any help would be greatly appreciated

Answer 1

Okay, you have:

(x-π/2)tan x

Which gives:

(x-π/2)sin x / cos x

Now [x→π/2]lim (x-π/2)sin x = 0 and [x→π/2]lim cos x = 0, so we may straightforwardly apply l'hopital's rule:

[x→π/2]lim (x-π/2)sin x / cos x = [x→π/2]lim (sin x + (x-π/2)cos x) / (- sin x).

(x-π/2)cos x clearly becomes zero as x→π/2, thus we have sin (π/2)/-sin (π/2), which is 1/-1, which is -1. Therefore:

[x→π/2]lim (x-π/2)tan x = -1

Answer 2

OK, just remeber there are 2pi radians in a circle, so pi radians is 180 degrees and pi/2 radians is 90 degrees.

As x approaches pi/2, (x-pi/2) will approach 0.
As x approaches pi/2, tanx will approach infinity because tan90 is infinity.

So, since 0 multiplied by anything is 0, the limit is 0.

To test this, use your calculator.
Display pi and subtract something like 0.0000000001.
use the resulting number as x .
You will find the answer very very close to 0 and -ve.

Answer 3

This is 0 times infinity.
change it to infinity over infinity.
Then find derivative of numerator and denominator

hint: 40+50=90
sin40=cos(90-40)

Let us begin to solve First let us change the way

sin(x-pi/2)=cos x
cos(x-pi/2)=sinx
tanx=sinx/cosx=cos(x-pi/2) over sin(x-pi/2)


lim (x-pi/2) over sin(x-pi/2)=0/0 So it is known as 1

because limit of sin/x=1

lim 1. cos(x-pi/2)= 1.cos 0=1.1=1

Answer 4

the result is -1

Answer 5

limit x>pi/2 (x-pi/2)[sinx/sin(pi/2-x)]
limit x>pi/2 -sinx/{sin(x-pi/2)/(x-pi/2)}
=-sinpi/2=-1