Posted by Liz on Wednesday, June 27, 2012 at 1:07pm.
**Note**
Please use the Epsilon-Delta Definition.
1)
f(x) = 2 x + 5
The limit for x to 3 is 11, assuming the function is continuous. To prove this, consider the open interval
I = (11-epsilon, 11+epsilon)
for some arbitrary epsilon
We want to find a delta > 0 such that the function f maps the open interval
J =(3-delta, 3+delta)
into I. That would then prove that for any arbitry epsilon>0 such a delta indeed exists. This means that no matter how small you take epsilon, there is always an interval containing x = 3 that will be mapped by f into the interval of size 2 epsilon containing 11.
If we solve:
2 x + 5 = y
for x, we get:
x = (y-5)/2
If we take y = 11 +/- epsilon, we get:
x = 3 +/- epsilon/2
So, we can take delta = epsilon/2
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