Limits of Piecewise Functions...1

In order for the limit to exists, the following must take place:

lim (x + 1) as x tends 0 from the left = lim (x^2 + 1) as x tends to 0 from the right.

Let x = 0.

(x + 1) = (x^2 + 1)

(0 + 1) = ((0)^2 + 1)

1 = 1

So, I can now say that the limit of f(x) as x tends to 0 is 1.

You say?

If this is right, I post 5 more on my next day off.

 

no interesting limit points found
limit at x=0 is undefined, though left and right limits exist.

 

It's not right?

 

Please explain this! Why is there no limit at x= 0? It looks to me like the limit at x= 0 is 1 as nycmathguy said.

 

at x=0 , the left lim is -1 and the right limit is 1; so, different

Existence of a limit means that it has one definite and finite, real value ( in one dimensional case). If the left limit and right limit are different, then they are two different values and so the ""one value " criteria is violated. The left hand limit as well as the right hand limit of the function should be the same if the function is to have a limit.

 

True but this is a piecewise function. Why is my setting wrong?

 

you did this:
Let x = 0.

(x + 1) = (x^2 + 1)

(0 + 1) = ((0)^2 + 1)

1 = 1
just proof that both equations of piecewise function are equal IF x=0

but, conditions are
for x+1 given that x<=0
for x^2+1 given that x>0

 

I found a video online with the same problem. If I am wrong, so is the math teacher on YouTube.

 

try this one

https://www.desmos.com/calculator/9y8itvyshy

or copy this and past to wolfram alpha

lim(Piecewise[{{x+1, x<=0}, {x^2+1, x>0}}] )

 

Is the math teacher on YouTube wrong?

 

he is right about existence of the left and right limit that are same
but limit at x=0 is undefined, though left and right limits exist

 

I will post a few limits of piecewise functions on my next set of days off.