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Reason for why limits not existing : In the previous section we saw two limits that did not.
We saw that
did not exist since the function did not settle down to a single value as t approached t = 0 . The closer to t = 0 we moved the more passionately the function oscillated & in order for a limit to exist the function have to settle down to a single value.
However we saw that did not present not since the function didn't settle down to a single number as we moved in towards t = 0 , but rather then because it settled into two distinct numbers based on which side of t = 0 we were on.
The problem was that, as we approached t =0 , the function was moving in towards different numbers on each of the side.
Given f (x) = - x 2 + 6 x -11 determine each of the following. (a) f ( 2) (b) f ( -10) (c) f (t ) Solution (a) f ( 2) = - ( 2) 2 + 6(2) -11 = -3 (
Graph y = cos (x) Solution: There actually isn't a whole lot to this one. Given the graph for -4 ? ≤ x ≤ 4 ? . Note that we can put all values of x in cosine (that wo
Use L''hopital''s rule since lim X-->0 1-cos(x)/1-cos(4x) is in the indeterminate form 0/0 when we apply the limt so by l''hoptital''s rule differentiate the numerator and den
How should shoppers''stop develop its demand forecasts?
prove that cos(a)/1-sin(a)=tan(45+A/2)
NULL/ VOID/ EMPTY SET A set which has no element is known as the null set or empty set and is indicated by f (phi). The number of elements of a set A is indicated as n (A) and
1.A=the set of whole numbers less tan 4 ? 2.B=the set of prime numbers less than 19 ? 3.C=the set of first three days of week?
Expand (1- 1/2x -x^2)^9
Simpson's Rule - Approximating Definite Integrals This is the last method we're going to take a look at and in this case we will once again divide up the interval [a, b] int
Polynomials in two variables Let's take a look at polynomials in two variables. Polynomials in two variables are algebraic expressions containing terms in the form ax n y m
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