let f(x)= ((3x-1)|x-1|) / (x-1) for x !=0
(a) sketch the graph of f(x)
(b)evaluate lim x->1+ f(x)
(c) evaluate lim x->1- f(x)
(d)evaluate lim x->1 f(x)

This behaves linearly in x except when x=1, where there is a jump discontinuity. This is because |x-1|/(x-1) is a constant equal to +1
when x>1, and a constant equal to -1 when x<1.


For x>1 we have:

f(x) = 3x-1

and for x<1, we have:

f(x) = -3x+1,

with a jump discontiuity where f changes grom -2 to the left of x=1
to ~2 to the right of x=1.

Now I have a sketch of this but the upload function appears to be
brocken

If the diagram had uploaded it would be clear that the limit as x goes to
1 from above is the limit as x goes to 1 of 3x-1 which is equal to 2.

If the diagram had uploaded it would also be clear that the limit as x goes to
1 from below is the limit as x goes to 1 of -3x+1 which is equal to -2.

As the two previous limits are not equal there is no such limit.

RonL

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Giordano Bruno

Hi,

I'll try to upload the graph. It's working!

Then here is the graph of your function:

Attached Thumbnails