- #1
nycmathdad
- 74
- 0
Use the epsilon-delta method to show that the limit is 3/2 for the given function.
lim (1 + 2x)/(3 - x) = 3/2
x-->1
I want to find a delta so that | x - 1| < delta implies |f(x) - L| < epsilon.
| (1 + 2x)/(3 - x) - (3/2) | < epsilon
-epsilon < (1 + 2x)/(3 - x) - 3/2 < epsilon
I now add 3/2 to all terms.
(3/2) - epsilon < (1 + 2x)/(3 - x) < (3/2) + epsilonStuck here...
lim (1 + 2x)/(3 - x) = 3/2
x-->1
I want to find a delta so that | x - 1| < delta implies |f(x) - L| < epsilon.
| (1 + 2x)/(3 - x) - (3/2) | < epsilon
-epsilon < (1 + 2x)/(3 - x) - 3/2 < epsilon
I now add 3/2 to all terms.
(3/2) - epsilon < (1 + 2x)/(3 - x) < (3/2) + epsilonStuck here...