1. ## discontinuity

f(x)=2sinx-cosx
-----------
1+tanx

In how many points in the interval [0,2pi] is the function above discontinious ?

I say 3 points book says 2.

pi/2:tangent is undefined
tan135:denominator becomes 0
tan315: denominator becomes 0.

2. ## Re: discontinuity

The function is continuous at $x= \frac{\pi}{2}$. Look at the limits as $x \rightarrow \frac{\pi}{2}$ from the left and from the right.

3. ## Re: discontinuity

-infinite from the right and +infinite from the left

4. ## Re: discontinuity

Then what can you say about limit of $f(x)$ as $x \rightarrow \frac{\pi}{2}$

5. ## Re: discontinuity

It has no limit.

6. ## Re: discontinuity

It tends to zero from both sides...

From the left ($\pi/2$), $\tan x$ tends to $+ \infty$. From the right, $\tan x$ tends to $-\infty$.

So $$\lim_{x \rightarrow \frac{\pi}{2}^-} f(x)= \frac{2}{1+\infty}=0$$
and
$$\lim_{x \rightarrow \frac{\pi}{2}^+} f(x)= \frac{2}{1-\infty}=0$$

So from this we can say that $$\lim_{x \rightarrow \frac{\pi}{2}} f(x) =0$$.

Now a function is continuous at a point $a$ if $$\lim_{x \rightarrow a} f(x) = f(a)$$

Can you re-write $f(x)$ so that you can evaluate it at $x=\pi /2$? Hint: Use $\tan x = \frac{ \sin x}{\cos x}$ and write $1+ \tan x = \frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}$ .

7. ## Re: discontinuity

Kastamonu, Ahri is offering a different function that is continuous at $\frac{\pi}{2}$. The function you are given is undefined at that point, so you are correct that is it discontinuous. The definition of continuous requires the function be defined AND that the limit from the left and from the right be equal to the value defined.

There will be four points of discontinuity on the given interval.

$\dfrac{\pi}{2},\dfrac{3\pi}{4},\dfrac{3\pi}{2}, \dfrac{7\pi}{4}$

8. ## Re: discontinuity

Originally Posted by SlipEternal
Kastamonu, Ahri is offering a different function that is continuous at $\frac{\pi}{2}$. The function you are given is undefined at that point, so you are correct that is it discontinuous. The definition of continuous requires the function be defined AND that the limit from the left and from the right be equal to the value defined.

There will be three points of discontinuity on the given interval.
I was going on $$f\left(x\right)=\frac{2\sin x - \cos x}{1+ \tan x}=\frac{2\sin x - \cos x}{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}=\frac{2\sin x - \cos x}{\frac{1}{\cos x}\left(\cos x + \sin x\right)}=\frac{\cos x \left(2\sin x - \cos x\right)}{\cos x + \sin x}$$

which can be evaluated at $x= \frac{\pi}{2}$

9. ## Re: discontinuity

Originally Posted by Ahri
I was going on $$f\left(x\right)=\frac{2\sin x - \cos x}{1+ \tan x}=\frac{2\sin x - \cos x}{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}=\frac{2\sin x - \cos x}{\frac{1}{\cos x}\left(\cos x + \sin x\right)}=\frac{\cos x \left(2\sin x - \cos x\right)}{\cos x + \sin x}$$

which can be evaluated at $x= \frac{\pi}{2}$
That would be like saying the rational function $\dfrac{x^2}{x}$ can be evaluated at 0.

10. ## Re: discontinuity

It's been a while since I've studied these things, apologies then .

11. ## Re: discontinuity

f(x)=(2sinx-cosx)/(1+tanx)
The function is like that.
tan135:denominator becomes 0
tan315: denominator becomes 0.
At these points it is discontinious
But I am not sure about 90 and 270. Tangent is undefined at these points.
According to the book I am wrong but Ahri is right.

12. ## Re: discontinuity

Originally Posted by kastamonu
f(x)=(2sinx-cosx)/(1+tanx)
The function is like that.
tan135:denominator becomes 0
tan315: denominator becomes 0.
At these points it is discontinious
But I am not sure about 90 and 270. Tangent is undefined at these points.
According to the book I am wrong but Ahri is right.
What is the definition of continuity according to your book?

13. ## Re: discontinuity

f:A--->R
a is element of A
There is a limit when x goes to a and this limit is equal to f(a).

If we put our finger on one end of the function and go to the other end without removing our finger then we can say that the function is continious.

14. ## Re: discontinuity

Perfect, since the function does not exist, it cannot be continuous when tangent does not exist.

15. ## Re: discontinuity

Sorry if I confused you Kastamonu, I'll be more careful next time .

Page 1 of 2 12 Last