calculus problem?

Question

this is a limit problem. The problem is:
lim (pi/2 - x)tanx
x-->pi/2
i found this problem on a book and it starts like this:
first step,x=pi/2+h and then x-->pi/2 and h--->0
But i don't understand why "x=pi/2+h" is written?how "h" comes?
please tell me reasons.

Answer 1

Perhaps it is easier to rewrite it in terms of sines and cosines?

\lim_{x \to pi/2} (pi/2 - x) tan(x)=
\lim_{x \to pi/2} (pi/2 - x) sin(x)/cos(x)=
\lim_{x \to pi/2} (pi/2 - x)/cos (x)* \lim_{x \to pi/2} sin(x)=
\lim_{x \to pi/2} (pi/2 - x)/cos (x)=1

From line 3 to line 4 I have used that sin(pi/2)=1.

\lim_{x \to pi/2} (pi/2 - x)/cos (x)=1 because
\lim_{x \to pi/2) cos(x)/(pi/2-x)=
-\lim_{x \to pi/2) (cos(x)-cos(pi/2))/(x-pi/2)=
the negative of the value of the derivative of cos(x) at pi/2:

d/dx (cos(x))=-sin(x). At pi/2, -sin(pi/2)=-1.

I believe the hint is trying to get you to see the last line in terms of the f'(x)=\lim_{h \to 0}(f(x+h)-f(x))/h, provided the limit exists version of the definition of the derivative.

Answer 2

The reason for the "h" is that you are examining the expression for values of x very close to pi/2. h is supposed to be a small number. The limit x-->pi/2 corresponds to h-->0. The ultimate question is: as h-->0 and therefore x-->pi/2, is pi/2-x shrinking faster than tan(x) is growing?

The "h" may also allow you to use some trig formuls to rewrite
tan(x) close to x=pi/2 in a more convenient form.

Answer 3

The limit is zero.

You can rewrite the above expression as:

(pi/2)tanx - xtanx

Now consider what happens to tanx as x--> pi/2. It becomes infinitely large. So both above terms become infinitely large. What is the difference between two infinitely large terms?
Easy: 0.

In the case where x--> pi from the other side, the one term becomes positive so that the result remains the same.

Answer 4

I believe you can just take the limit as tanx approaches pi/2, which is 0 because of the squeeze theorem. It's been awhile since I have done this stuff. Good luck.

Answer 5

when x->pi/2 x-pi/2 tends to zero. So there is a number x-pi/2 we call it h. x->pi/2 means h->0 that is all