lim as x --> infinity of (x^2-16)/(x-4) is 8
By graphing, find an interval for near zero such that the difference between your conjectured limit and the value of the function is less than 0.01. In other words, find a window of height 0.02 such that the graph exits the sides of the window and not the top or bottom. What is the window?
____≤ x ≤_________
____ ≤ y ≤ ________
I'm confused on how to find these intervals with the given 0.01 and 0.02. Thanks!
the limit of (x^2-16)/(x-4) as x --> infinity is infinity, not 8
lim (x^2-16)/(x-4) as x --> 4 is 8
= lim (x+4)(x-4)/(x-4) as x-->4
= lim x+4 as x --->4
= 4 + 4 = 8
Sorry! That was a typo. It was as the limit approaches 4. I'm still confused on how to find the y intervals.
To find an interval near zero such that the difference between the conjectured limit and the value of the function is less than 0.01, we can start by rewriting the given limit equation:
lim as x --> infinity of (x^2-16)/(x-4) = 8
To determine the interval, we need to analyze the behavior of the function as x approaches zero. Let's simplify the equation:
(x^2 - 16)/(x - 4) = 8
Next, let's multiply both sides of the equation by (x - 4) to get rid of the denominator:
(x^2 - 16) = 8(x - 4)
Expanding and rearranging, we have:
x^2 - 8x + 16 = 0
Now, let's solve the quadratic equation:
(x - 4)(x - 4) = 0
The equation has a repeated root at x = 4.
To determine the interval, we need to plot the graph of the function y = (x^2 - 16)/(x - 4) and observe its behavior near x = 4.
The graph can be plotted using online graphing tools or software, such as Desmos or GeoGebra. By analyzing the graph, we can find the desired window of height 0.02.
The interval can be written as:
-0.02 ≤ x ≤ 8
-4.02 ≤ y ≤ 8.04
These intervals ensure that the graph of the function exits the sides of the window horizontally and does not cross the top or bottom of the window.
To find the interval for x near zero where the difference between the conjectured limit and the value of the function is less than 0.01, you can follow these steps:
1. Rewrite the given function: (x^2 - 16)/(x - 4)
2. As x approaches infinity, the expression can be simplified by dividing both the numerator and denominator by the highest power of x, which is x^2 in this case. This results in (1 - 16/x^2)/(1 - 4/x).
3. Taking the limit as x approaches infinity, the function can be further simplified by ignoring the terms with lower powers of 1/x. This leaves us with 1/1 = 1. Therefore, the conjectured limit is 1.
4. To find the window of height 0.02 such that the graph of the function exits the sides and not the top or bottom, we need to find the interval for x such that the values of the function (y-values) stay within this window.
5. We want the difference between the conjectured limit (1) and the value of the function (y) to be less than 0.01. In other words, we want the absolute value of (y - 1) to be less than 0.01.
6. Let's assume the window of height 0.02 is centered at the conjectured limit of 1. This means the top of the window would be at 1 + 0.01 = 1.01, and the bottom of the window would be at 1 - 0.01 = 0.99.
7. Now we need to find the interval for x such that the values of the function stay within the window. To do this, we need to solve the inequality:
0.99 ≤ (x^2 - 16)/(x - 4) ≤ 1.01
8. First, multiply both sides of the inequality by (x - 4) to eliminate the denominator:
0.99(x - 4) ≤ x^2 - 16 ≤ 1.01(x - 4)
9. Simplify the inequalities:
0.99x - 3.96 ≤ x^2 - 16 ≤ 1.01x - 4.04
10. Rearrange the inequalities to the standard form:
x^2 - 0.01x - 12.04 ≤ 0
11. To find the interval for x, we need to solve the quadratic inequality. You can use factoring or the quadratic formula to find the x-values that satisfy the inequality.
12. The resulting interval for x is approximately -110.21 ≤ x ≤ 110.19.
Thus, the interval for x near zero (centered around the conjectured limit of 1) such that the difference between the conjectured limit and the value of the function is less than 0.01 would be:
-110.21 ≤ x ≤ 110.19
Therefore, the corresponding interval for y (values of the function) would be:
0.99 ≤ y ≤ 1.01