Question

Find all values of x at which the function g(x) is Discontinuous.

Answer 1

\[\LARGE g(x)=ln(1+sinx)\]

Answer 2

under what conditions that a function is discontinuous?

Answer 3

If it has a "cusp" or break?

Answer 4

Do you agree that \(\sin\) is a continuous function? And if yes, do you agree that \(\ln\) is a continuous function on the positive domain \((0, + \infty) \) ?

Answer 5

if so, what does that tell you about the composition of the two functions?

Answer 6

Also let me highlight here, having a "cusp" (I am not fluent in english, but I did translate it, meaning a bulge/edge I suppose?) does not mean that the function is discontinuous at this point. Check the Waerden function for example, or as a simple exercise check for the function \(f(x)=|x|\) this function is continuous in \(\mathbb{R}\), yes also on \(0\). However it is not differentiable on \(x=0\)

Answer 7

here is a link to the infamous Waerden function http://www.encyclopediaofmath.org/index.php/Non-differentiable_function if you care to have a look.

Answer 8

I can see, graphically, that it is discontinuous at pi/2, but anyway to show it algebraically?
And cusp as in a sharp corner.. like in the |x|, at 0.

Answer 9

Are you familiar with the epsilon delta criteria?

Answer 10

I guess you want to analyze this function on the real domain, then you could try for instance to show if the following limit exists:
\[\large \lim_{ x \to - \frac{\pi}{2}} \ln ( 1 + \sin x) \]

Answer 11

cusp or sharp edge is not a criterion to make a function discontinuous
it may mean the existence of a piecewise-defined function though. unless the line is broken or at some point a hole in the graph, it is continuous.

Answer 12

For the epsilon delta criteria, try to negate the following statement:
\[\large \forall \epsilon > 0, \exists \delta > 0, \forall x \in \mathbb{R} :( |x-x_0|< \delta \rightarrow |f(x)-f(x_0) < \epsilon ) \]

Answer 13

a bit of an overkill though.

Answer 14

yeah. an absolute value function is continuous, though limit DNE

Answer 15

For the sake of the argument check between approaching your limit from the left or from the right to show divergence in your case (to plus and minus infinity).