# Calculus

posted by on .

I have two similar problems that I need help completing. Please show all your work.

Question: Find the limit L. Then use the å-ä definition to prove the limit is L.

1. lim (2x+5)
x->3
2. lim 3
x->6
Thank you for your anticipated help!!

• Calculus - ,

**Note**

• Calculus - ,

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