finding one-sided lim + question

[orir]

How do i find one sided limit of a function, in case i don't have its plot (or a view as a group)?
I mean, in order to find rather a certain limit exists i need to find its existence in both sides, and sometimes it's different values - but if i simply put the value in the function i get the same result...

and, another question:
Why are the functions of arcsin(sinx) and sin(arcsinx) not the same (x)?

 

[daon2]

To find the "+" or right-hand limit, you assume x is larger than the value in question. You cannot always plug in the value; a function must be right-continuous for plugging in to work.

Example: What is the limit as x approaches 2 from the right of (x+2)/(x^2-4)? You cannot plug in 2, so you assume x>2, then you may reduce your function to 1/(x-2). There you can see your function has a vertical asymptote at x=2, and if x>2, the functions value is positive. Hence the limit is +infinity.

For your second question, sin(arcsin(x)) does always equal x. But arcsin(sin(x)) will only equal x if x belongs to a certain interval. Can you determine which interval that is?

 

[orir]

But..

My question is more formal - why these two functions are not mathmematically equal? Aren't reverse functions work both ways?!

 

[JeffM]

My question is more formal - why these two functions are not mathmematically equal? Aren't reverse functions work both ways?!

Your questions are not formal; they are vague. What two functions? By "reverse" functions, do you mean "inverse" functions. If so, what does that have to do with limits?

How do i find one sided limit of a function ...
I mean, in order to find rather a certain limit exists i need to find its existence in both sides, and sometimes it's different values - but if i simply put the value in the function i get the same result...

Are you saying that \(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = f(a)\) for every function at every a? That is false.

\(\displaystyle x< 3 \implies f(x) = - x.\)

\(\displaystyle x = 3 \implies f(x) = 0.\)

\(\displaystyle x > 3 \implies f(x) = 2x.\)

In this case, the right limit exists as x approaches 3 from above; left limit exists as x approaches 3 from below; function exists at x = 3, and all three values are different. Sometimes right limit and left limit both exist and are the same. Sometimes both exist and are different. Sometimes one exists and not the other.

 

[lookagain]

My question is more formal - why these two functions are not mathmematically equal?
Aren't reverse functions work both ways?!

orir,

look at an additional example:


\(\displaystyle Compare \ \ \sqrt{x^2} \ \ with \ \ (\sqrt{x})^2.\)


For non-negative values of x, these give the same value.


When x equals a negative value, \(\displaystyle \ x^2 \ \) is necessarily positive.

Then the primary square root of that value is positive.


Look at \(\displaystyle (\sqrt{x})^2.\)

It equals \(\displaystyle \ (\sqrt{x})(\sqrt{x}).\)


The rule for this is that this product equals x, regardless of the sign of x.


As a particular example of this, look at \(\displaystyle \ (\sqrt{-9})^2.\)


\(\displaystyle \ (\sqrt{-9})^2 \ = \)


\(\displaystyle \ (\sqrt{-9})(\sqrt{-9}) \ = \)


-9


- - - - - - - - \(\displaystyle \ \ or, \ \ put \ \ another \ \ way \ \ \) - - - - - - - -


\(\displaystyle \ (\sqrt{-9})^2 \ = \)


\(\displaystyle (3i)^2 \ = \)


\(\displaystyle 9i^2 \ = \)


\(\displaystyle 9(-1) \ = \)


-9

 

[daon2]

My question is more formal - why these two functions are not mathmematically equal? Aren't reverse functions work both ways?!

Certainly they are not the same. Functions f and g are the same if they have the same domain, the same codomain, and f(x)=g(x) for all x in the domain.

f(x)=sin(arcsin(x)) takes as input an x-value in [-1,1] and gives an output in [-1,1]. This function is one-to-one.

g(x)=arcsin(sin(x)) takes any real number as input and outputs a value between [-pi/2,pi/2]. This function is many-to-one.

They are very different functions. The inverse-sine function is only defined for sine on an interval of length pi, [-pi/2, pi/2], because if it were any longer sine would fail to be one-to-one.

 

[HallsofIvy]

How do i find one sided limit of a function, in case i don't have its plot (or a view as a group)?
I mean, in order to find rather a certain limit exists i need to find its existence in both sides, and sometimes it's different values - but if i simply put the value in the function i get the same result...

Many beginning Calculus students unfortunately get the idea that "limit of f at x= a" is just a fancy way of saying "the value of f at a". That's not true. What is true is that continuous functions have that property. (That is the definition of "continuous".) In fact, "almost all" functions are NOT continuous but we use continuous functions a lot because they are so nice. You take the two one sided by looking only at the formula for the function on that side. For example, if f(x)= x- 1 for x<0, f(0)= 2, and \(\displaystyle f(x)= x^2+ 1\) for x> 0, to find \(\displaystyle \lim_{x\to 0^-} f(x)\) \(\displaystyle \lim_{x\to 0} x- 1\). That's continuous at x= 0 (all polynomials are continuous for all x) that limit is 0- 1= -1. To find \(\displaystyle \lim_{x\to 0^+} f(x)\), look at \(\displaystyle \lim_{x\to 0} x^2+ 1\). Again that is continuous so the limit is \(\displaystyle 0^2+ 1= 0\). Since the two one sided limits are differernt, the limit itself, \(\displaystyle \lim_{x\to 0} f(x)\) does not exist. And, of course, then the function is not continuous at x= 0.

and, another question:
Why are the functions of arcsin(sinx) and sin(arcsinx) not the same (x)?

IF f is "one-to-one" (f(x)= f(y) only if x= y), then \(\displaystyle f^{-1}(f(x))= f(f^{-1}(x)= x\). If f is not one-to-one, then, strictly speaking, f does not have an inverse. But we can, and typically do, restrict the domain of f and define an inverse for that. For example, \(\displaystyle f(x)= x^2\) is not one-to-one because f(2)= f(-2). What we do is restrict f to only non-negative numbers and its inverse is "the positive number whose square is equal to that"- \(\displaystyle \sqrt{x}\). it is true that \(\displaystyle f(f^{-1}(x))= x\) for all x in the domain of square root. However, \(\displaystyle x^2\) has domain all real numbers and if, say, x= -2, \(\displaystyle f^{-1}(f(2))= f^{-1}(4)= 2\), not -2.

The same thing happens for sine- only worse! sine is periodic with period \(\displaystyle 2\pi\) so, to have an inverse, we need to restrict the range of arcsine to \(\displaystyle -\pi\) to \(\displaystyle \pi\). So for, say \(\displaystyle x= 2\pi\), \(\displaystyle arcsin(sin(x)= arcsine(0)= 0\), not \(\displaystyle 2\pi\).