lim
x---> 2 4x+1/3x-4
illustrate definition 2 by finding values of delta that correspond to epsilon=0.5 and epsilon= 0.1
I having trouble setting this up.
do i set up two different functions then divide?
0.2898 that's the answer i got....
clearly the limit is (8+1)/(6-4) = 9/2
Not sure what your definition 2 is, but if you need to find δ such that |f(x+δ)-9/2| < 0.5 then that means
|(4(2+δ)+1)/(3(2+δ)-4)-9/2| < 0.5
Work your way through that and you will find that δ < 0.125
So, if you pick a value of x within 1/8 of 2, f(x) will be within 1/2 of 9/2.
Or, more mathematically,
if |x-2| < 1/8, |f(x)-9/2| < 1/2
To illustrate the definition of the limit, we need to find suitable values of delta that correspond to given epsilon values. Let's consider the function:
f(x) = (4x + 1) / (3x - 4)
We want to find the values of delta such that for any epsilon greater than zero, there exists a delta greater than zero such that if 0 < | x - 2 | < delta, then | f(x) - L | < epsilon.
First, let's consider epsilon = 0.5:
We need to find a delta such that if 0 < | x - 2 | < delta, then | f(x) - L | < 0.5.
To do this, we can manipulate the expression for f(x) by multiplying the numerator and denominator by the conjugate of the denominator:
f(x) = (4x + 1) / (3x - 4)
= (4x + 1) * (3x + 4) / (3x - 4) * (3x + 4)
= (12x^2 + 16x + 3x + 4) / (9x^2 - 16)
Now we can simplify f(x):
f(x) = (12x^2 + 19x + 4) / (9x^2 - 16)
To find delta corresponding to epsilon = 0.5, we need to find a range of x values around 2 where the function f(x) stays within 0.5 of the limit L. In other words, we want to find the range of x values where:
| f(x) - L | < 0.5
Let's set L = 0.2898, as you mentioned, and solve for f(x):
| f(x) - 0.2898 | < 0.5
Now, you can set up the inequality | f(x) - 0.2898 | < 0.5 and solve algebraically for the range of x values.
For epsilon = 0.1, you would repeat the same process, setting up the inequality | f(x) - L | < 0.1. With the given function, you might obtain a different value for L. Follow the same steps as above to solve for the corresponding delta.
Remember that finding suitable values for delta requires a careful analysis of the function and utilizing algebraic manipulations to simplify and isolate the variable of interest.