# Calc

posted by Brittany

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!

1. Reiny

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

2. Josie

Sorry! That was a typo. It was as the limit approaches 4. I'm still confused on how to find the y intervals.

## Similar Questions

1. ### calc limit

lim (2^x - 3^-x)/2^x + 3^-x x-> infinity Thanks. Do you mean (2^x - 3^-x)/(2^x + 3^-x) or [(2^x - 3^-x)/2^x] + 3^-x ?
2. ### precalcus

can someone explain how to evaluate limits?
3. ### Calculus - ratio test

infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] | = lim (n->infinity) | ((e^n)(e^1)(n!)) …
4. ### calculus - ratio test

infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] | = lim (n->infinity) | ((e^n)(e^1)(n!)) …
5. ### Calc. Limits

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity?
6. ### Calculus

Consider the function f(x)=sin(5x)/x. (a) Fill in the following table of values for f(x): x= -0.1 -0.01 -0.001 -0.0001 0.0001 0.001 0.01 0.1 f(x)= ( I need the values of f(x) for each x) (b) Based on your table of values, what would …
7. ### Calculus

Consider the function f(x)=(4^x−1)/x. (a) Fill in the following table of values for f(x): x= -0.1 -0.01 -0.001 -0.0001 0.0001 0.001 0.01 0.1 f(x)= I the the falues of f(x) for each interval... (b) Based on your table of values, …
8. ### Calculus

Consider the function f(x)=(4^x−1)/x. (a) Fill in the following table of values for f(x): x= -0.1 -0.01 -0.001 -0.0001 0.0001 0.001 0.01 0.1 f(x)= I the the falues of f(x) for each interval... (b) Based on your table of values, …
9. ### calc

Use L’Hopital’s rule to find the limit of this sequence (n^100)/(e^n) ...If you do L'Hop. Rule it would take forever, right?
10. ### Calculus

Consider the function f(x)=(5^x−1)/x. A) Fill in the table values for f(x): x= -0.1, -0.01, -0.001, -0.0001, 0.0001, 0.001, 0.01, 0.1 f(x)= 1.4866, 1.5866, 1.6081, 1.6093, 1.6096, 1.6107, 1.6225, 1.7462 B) Based on the table …

More Similar Questions