Question

for x not = pi/2 is continous at x=pi/2
find f(pi/2)

Answer 1

\[f(x)=\sqrt{2 }-\sqrt{1+\sin x}\div \cos ^2x\]

Answer 2

@ikram002p

Answer 3

but its not continues at pi/2 ,
do u mean find limit?

Answer 4

mmm yes

Answer 5

@ikram002p

Answer 6

\(\Huge \frac {\sqrt 2 -\sqrt {1+\sin x} }{\cos^2 x}\)

hint :-
use LH rule
do you know what it is ?

Answer 7

no

Answer 8

@ikram002p

Answer 9

ohh :)
so if the limit was 0/0 or somthing/0
then use LH rule
brb to explain

Answer 10

@kropot72 pls help me

Answer 11

@IMStuck

Answer 12

so to use LH rule to find the limits ,if u have a function of numerator and denominator
u only have to find derivative of both numerator and denominator
and then try to find the limit , its like this

\(lim \dfrac {g(x)}{h(x)}=lim \dfrac {g'(x)}{h'(x)}\)

\(lim \dfrac {g'(x)}{h'(x)}=lim \dfrac {g''(x)}{h''(x)}\)

Answer 13

\[(\cos x \sin 2x)\div 2\sqrt{1+\sin x}\]

Answer 14

@dan815 could you continue here plz ?
@rvc is my cute friend she is nice
but i have to go for a while :'(

Answer 15

@dan815

Answer 16

okya so lets see if this is still indeterminate form

Answer 17

lhopital can be reapplied as long as its indeterminate form, or now

Answer 18

idk that

Answer 19

give me the rule pls

Answer 20

okay so when its 0/0, at the given limit or +/- inf/inf at the given limit, we can differentiate the top and bottom of the fraction separately and it will equal the same thing

Answer 21

lets continue where u left off

Answer 22

\[\Huge \frac {\sqrt 2 -\sqrt {1+\sin x} }{\cos^2 x}\]

Answer 23

okay

Answer 24

at f(pi/2) we have 0/0 so lets apply LH to this

Answer 25

i have written the derivate

Answer 26

\[\Huge \frac {\sqrt 2 -\sqrt {1+\sin x} }{\cos^2 x}\]
applying LH gives
\[\frac{1/2\sqrt {(1+sinx)}*cosx}{2cosx*-sinx}\]

Answer 27

have i done right?

Answer 28

nope because still indeterminate when evaluated at pi/2 so we can apply again

Answer 29

oh wait no need :)

Answer 30

u can cancel the cos and then it can be evaluated

Answer 31

-2cosx*sinx=-sin2x

Answer 32

\[\frac{1/2\sqrt {(1+sinx)}}{2*-sinx}\] \[evaluate~ at~\pi/2 \]

Answer 33

\[\sqrt2/-4\]

Answer 34

we can write sin2x

Answer 35

?

Answer 36

@dan815

Answer 37

that is not good to do because u will have a 0 in the denominator

Answer 38

hold on

Answer 39

@dan815

Answer 40

okay yep it is right i believe

Answer 41

wait unless there is no negative

Answer 42

http://www.wolframalpha.com/input/?i=y%3D%28sqrt+2+-sqrt+%7B1%2Bsin+%28x%29%7D%29+%2Fcos%5E2+%28x%29

Answer 43

look at this

Answer 44

oh ya oops forgot a negative on top

Answer 45

it should be \[\sqrt2/4\]