find lim of f(x) where f(x) is a piece wise function As follows f(x) = | x | + 1, x 0 where x is approaching a ( x - a ) for what value(s) of a does lim f(x ) exits? x-a

Question

find lim of f(x) where f(x) is a piece wise function As follows f(x) = | x | + 1, x < 0
                             0 , x = 0 
                            | x | - 1, x > 0 where x is approaching a ( x -> a ) 

for what value(s) of a does lim f(x ) exits?
                                             x->a

turingtest · Answer

answers are not given for free here, you must try to learn
so what is$$\lim_{x\to a^+}f(x)$$

turingtest · Answer

ok, so what is it ?
:)

myininaya · Answer

Oh I see what values does the limit exist for

myininaya · Answer

Well where is the function broken up at?

anonymous · Answer

f(x) = -x + 1 ,  x < 0 
         0         , x = 0 
         x - 1   , x > 0  where  x -> a      
question find the value(s) of a for what lim f(x) exits where x -> a

myininaya · Answer

Like lets think where the limit doesn't exist possibly at first?

myininaya · Answer

Yes I asked you a question
where is the function broken up at?

anonymous · Answer

this is the whole statement above in here i have sent u . i really don't know anything else. the above statement is the simplified version of the original statement

turingtest · Answer

the information you provided is enough to answer myininaya's Q
try to work with her please

myininaya · Answer

I'm trying to help you answer your question though. I understand the question. I'm asking can you observe for me where the function is broken at?

anonymous · Answer

yah functions is broken when a = 0 . m i right

myininaya · Answer

Yes now we should test that value to see if the limit exists or not
Find left and right limit for x=0

myininaya · Answer

$$\lim_{x \rightarrow 0^+}(|x|-1)$$
$$\lim_{x \rightarrow 0^-}(|x|+1)$$
Find  these limits and see if these are equal

myininaya · Answer

The reason I used |x|-1 for the right of 0 is because it says that we have |x|-1 for x>0
The reason I used |x|+1 for the left of 0 is because it says that we have |x|+1 for x<0

myininaya · Answer

If the above limits are equal then the limit exists
if the above limits are not equal then the limit does not exist

anonymous · Answer

i am reading ur lines above it would take some time for me to understand after reading all statements i will tell u what i have understood :) really myininaya u r so helpful

myininaya · Answer

It is just a matter of pluggin' in by the way to find those above limits since the |x|-1 and |x|+1 are both continuous at x=0

anonymous · Answer

u said that both are continuous at x = 0.      otherwise they are not continuous ?

myininaya · Answer

? no

myininaya · Answer

Where are you stuck at? The pluggin' part?

anonymous · Answer

myininaya.. i got everything thanks a lot really