1. Find the limit as x approaches infinity of the function sin 18x/10x
2. Find the limit as x approaches 6 of the function 3/(x-6)^2
3. Find the limit as y approaches 0 of the function (sin 11y)/13y or state that the limit does not exist.
4. Find the limit by substitution of the following: the limit as x approaches -2 of the function -8x, or state that the limit does not exist.
5. Find the first derivative of the y= (6 - x^2)(x^3 - 3x + 1) by a) applying the product rule and b) multiplying the factors to produce a sum of simpler terms to differentiate.
6. Find the first derivative by applying the product rule, and then find the first derivative by multiplying the factors to produce a sum of simpler terms to differentiate.
y= (7x^2 + 3)(8x - 5 + 9/x )
7. Find the derivative of the function y=x^10e^x
8. A body moves on a coordinate line such that it was a position s=f(t)=t^2 - 9t + 8 on the interval 0 is less than or equal to t which is less than or equal to 8, with s in meters and t in seconds. (Answers need to be in integer or simplified fraction form)
A) Find the body's displacement and average velocity for the given time interval.
B) Find the body's speed and acceleration at the endpoints of the interval.
C) When, if ever, during the interval does the body change direction?
Because range of sin 18x is always [1,-1].
-1/10x <= sin(18x)/10x <= 1/10x
As x -> infinity, 1/x -> 0 as well as 1/(10x) -> 0 and -1/(10x) -> 0
Therefore, by the sandwich rule,
as x-> infinity : sin(18x)/10x -> 0
Hence, answer is 0.
Lim x -> 6 3/(x-6)^2
Right hand limit:
x = h+6
Lim h->0 3/(h+6-6)^2 = Lim h->0 3/h^2 = infinity
Left hand limit:
x = 6 - h
Lim h->0- 3/(6-h -6)^2 = Lim h->0 3/(-h)^2
= Lim h->0 3/h^2
As RHL = LHL = infinity
Hence, limit exists and function tends to infinity as x->6. --Answer
Lim y->0 (sin(11y)/(13y))
Put 11y = x
As y -> 0 : 11y->0
Lim y->0 (sin(11y)/(13y)) = Lim y->0 (sin (11y) / (13*11y/11))
= Lim x->0 (sin (x) / (13*x/11))
= Lim x->0 (11/13) (sin (x) / x)
= (11/13)*1 [Because, Lim x->0 sin(x)/x = 1]
= 11/13 --Answer
Lim x->(-2) (-8x)
Right Hand Limit:
x = -2 + h
Lim h->0+ (-8(-2+h)) = Lim h->0 (16 - 8h)
= 16 - Lim h->0 (-8h)
= 16 - 0
Left Hand Limit:
Put x = -2 - h
Lim h->0- (-8 (-2 - h)) = Lim h->0 (16 + 8h)
= 16 + ...
A few problems of limit are solved. Also, a few problems of first derivative which use product rule, are solved. A derivative application problem from kinematics is also solved.