Post on 30-Aug-2018
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One-Sided Limits and Continuity
Section 2.5
We have already discussed the right hand limit , and the(x)limx→a+
f
left-hand limit . Let us see the formal definition of(x)limx→a−
f
continuity of a function at a point .(x) f x = a
Definition: A function is continuous at a number if the(x)f x = a
following conditions are satisfied.
1. is defined.(a)f
2. exists.(x)limx→a
f
3. (x) (a)limx→a
f = f
Geometrically this means the graph of the function does not have
a break over/at the point with .x = a
Q. At what points is the above function continuous.
Soln. .= − ,− , }x / { 8 2 6
If is not continuous at , then we say that is(x)f x = a (x)f
discontinuous at .x = a
is said to be continuous in an interval if is(x)f a, ][ b (x)f
continuous at every point in the given interval.
Q. Is the function (x) x if x =f = { + 2 / 2
{3 if x = 2 Continuous at .x = 2
Soln. f(2) i defined.
exists(x) limx → 2
f = 2 + 2 = 4
But .(x) = (2)limx → 2
f / f
So the function is discontinuous at x=2. But it is continuous at
all other values of x.
Q. Find all points at which the function is continuous.(x)f = x−2x −42
Soln. The domain of this function is . The function is notx = }{ / 2
continuous at , since it is not defined there. But it isx = 2
continuous at all points in the domain of the function.
Note: as functions. For example is not in(x) =f = x−2x −42 / x + 2 x = 2
the domain of , but it is in the domain of . But at all(x)f x + 2
values of different from 2, the two functions have the samex
value. The graphs of are respectively.,x−2x −42 x + 2
Remark: Polynomial functions, rational functions and power
functions are continuous at every point in their respective
domains.
Q. Find the points of discontinuity of the following functions.
+1, , 1, , 3x xx −2x+12x +2x+12 √2x + 1 1x2 2 + 4
Soln. 1}, {x }, }, 0}, }{ < − 21 { { {
Remark. Except for piecewise defined function, every other
function we will see in this course would be continuous at every
point in their respective domains.
In other words if is a number between and . Then theN (a)f (b)f
equation has a solution between and , when f(x) is a(x)f = N a b
continuous function.
Q: Show that the equation has a solution in .x3 + x − 5 = 0 0, ][ 2
Soln. Consider the function . Then , and(x)f = x3 + x − 5 (0) −f = 5
. Since is a continuous function, and ,(2)f = 5 (x)f − 5 < 0 < 5
there must be a between and , such that .c 0 2 (c)f = 0
Please note, IVT may fail, if the function is discontinuous.
Continuity is necessary for the IVT to hold for sure.
for this function the equation
has no solution.(x)f = 3
The Intermediate value theorem has the following corollary.
(Existence of zeroes of a continuous function) If is af
continuous function on a closed interval , and if anda, ][ b (a)f
have opposite signs, then there is at least one solution of(b)f
the equation in the interval .(x)f = 0 a, )( b
Geometrically this means, that if the graph of a continuous
function, starts below x-axis at , and goes above x-axis atx = a
, or vice-versa, then it must cross the x-axis at some x = b x = c
. This is the root of the equation . But this may failx = c (x)f = 0
if the function is discontinuous.
But if is not continuous. Then IVT fails.(x)f
Q. Show that the function . has a zero in the(x)f = x3 + x − 1
interval . Find an approximate value of this zero.0, )( 1
Soln. is a polynomial function, hence it is continuous(x)f
function. f(0)= -1 < 0 and f(1) = 1 > 0. So by the intermediate
value theorem, f(x) must have at the least one zero in (0,1) .
f(0) = -1, f(1) = 1 , f(x) has a zero in (0,1)
f(0.5) = -0.375, f(x) has a zero in (0.5,1)
f(0.75) = 0.1718 f(x) has a zero in (0.5,0.75)
f(0.625) = -0.13 f(x) has a zero in (0.625,0.75)
f(0.6875)=0.012 f(x) has a zero in (0.625,0.6875)
f(0.65625) = - 0.0611 f(x) has a zero in (0.65625,
0.6875)
f(0.671875) = - 0.02482 f(x) has a zero in (0.671875,0.6875)
f(0.6796875) = - 0.0063138 f(x) has a zero in (0.6796875,0.6875)
Hence 0.68359375 0.68 is a good approximation for a zero.≈
This method is called the Method of Bisection. To check f(0.68)
= - 0.005568 . Hence this is a zero upto 2 decimal places.
Q. At what values of , will the following function bek
continuous.
(x) if x =f = { x+3x −92 / − 3
k if x{ = − 3
Soln. This function is continuous at every value of x, except possibly at x=-3. At x=-3, . So the limit6lim
x→ − 3 x+3x −92 = lim
x→ − 3x − 3 = −
exists and the function is defined, so the first 2 conditions for
continuity at x = - 3 are satisfied. For the third condition to
also hold good at x = - 3, we must have .k = − 6
Consider the following graph.
Q. What is the slope of the “secant line” S joining the points (a,f(a)) and (a+h,f(a+h))
Soln. .(a+h) − af(a+h) − f(a) = h
f(a+h) − f(a)
Q. What is the slope of the tangent line” T at the point (a,f(a)) .
Soln. limh → 0 h
f(a+h) − f(a)
Another answer : limx → a x − a
f(x) − f(a)
Q. What is the slope of the tangent line to the graph of at(x)f
the point .x, (x))( f
Soln. limh → 0 h
f(x+h) − f(x)
Q. Find the slope of the tangent line at the point (x,f(x)) for the function .(x) x xf = 2 2 + 5
Soln. We have to find limh → 0 h
f(x+h)−f(x) = limh → 0 h
[2(x+h) +5(x+h)]−[2x +5x]2 2
We use the 4 step process :
Step 1 . (x ) (x ) (x ) (x xh) x h x h xh x hf + h = 2 + h 2 + 5 + h = 2 2 + h2 + 2 + 5 + 5 = 2 2 + 2 2 + 4 + 5 + 5
Step 2. (x ) (x) 2x h xh x h] 2x x]f + h − f = [ 2 + 2 2 + 4 + 5 + 5 − [ 2 + 5
x h xh x h x x h xh h= 2 2 + 2 2 + 4 + 5 + 5 − 2 2 − 5 = 2 2 + 4 + 5
(2h x )= h + 4 + 5
Step 3. h xhf(x+h)−f(x) = h
h(2h+4x+5) = 2 + 4 + 5
Step 4. Slope of tangent at (x,f(x)) is ( 2h x ) 4x lim
h → 0 hf(x+h) − f(x) = lim
h → 0+ 4 + 5 = + 5
Caution: In the above limit is like a constant while is thex h
variable going to .0
Finally : the slope of the tangent line to the graph of at the(x)f
point is .x, (x))( f (x) xf ′ = 4 + 5
Q. Find the slope of the tangent line at the point (1,7) to the graph of .(x) x xf = 2 2 + 5
Soln. Here x=1. But we know slope at (x,f(x)) is 4x+5. So the
slope of the tangent at (1,7) is (1) (1) .f ′ = 4 + 5 = 9
Definition. The derivative of a function is a function ,(x)f (x)f ′
whose value is equal to the slope of the tangent to the graph of
at the point .(x)f x, (x))( f
So .(x)f ′ = limh → 0 h
f(x+h) − f(x)
Q. Find if .(x)f ′ (x) x xf = 2 2 + 5
a. Find the slope of the tangent line to the graph of at(x)f
the point 3, 3).( 3
b. Find the equation of the tangent line at the point (3,33).
Soln. .(x) xf ′ = 4 + 5
a. Slope at (3,33) is .(3) 7f ′ = 1
b. The tangent line has slope 17 and passes through (3,33) so
using the point-slope form , it’s equation is (y-33) =
17(x-3) . the standard form is y=17x-18.
Q. If . Find .(x) f = 1x−2 (x)f ′
Soln. We have to find for limh → 0 h
f(x+h) − f(x) (x) .f = 1x−2
Step 1. (x )f + h = 1(x+h)−2 =
1x+h−2
Step 2. (x ) (x)f + h − f = 1x+h−2 −
1x−2 = (x+h−2)(x−2)
(x−2)−(x+h−2) = x−2−x−h+2(x+h−2)(x−2) =
−h(x+h−2)(x−2)
Step 3. hf(x+h)−f(x) = h
−h(x+h−2)(x−2) = −h
(x+h−2)(x−2)h =−1
(x+h−2)(x−2)
Step 4. . limh → 0 h
f(x+h)−f(x) = limh → 0
−1(x+h−2)(x−2) =
−1(x−2)2
(We get the above limit by plug-in-g in h=0)
So So (x) −f ′ = 1(x−2)2
Q. Find the equation of the tangent line to the graph of at the point (1,-1).(x) x )f = ( − 2 −1
Soln. The slope of the tangent at this point is . So the(1)f ′ = − 1
eqn of the tangent is .− ) (x 1)y − ( 1 = − 1 −
⇒ y + 1 = − x + 1 ⇒ y = − x
Definition. The average rate of change of a function on an(x)f
interval is defined to be a, ][ b b−af(b)−f(a)
The ( instantaneous ) rate of change of a function at , is(x)f x = a
defined to be equal to .(a)f ′
Q. For the function . Calculate the average rate of(t) f = 1t−2
change of the function over the interval [1,3]. Also find the
instantaneous rate of change at t=1.
Soln. Av rate of change over the interval [1,3] =
/2 .3−1f(3)−f(1) = 2
1−(−1) = 2 = 1
Instantaneous rate of change at t=1 = (1) . f ′ = −1(1−2)2 = − 1
Q. Let the position of a car at the time be given by thet
function (t) 2t t) meters.s = ( 2 + 5
a. Find the velocity and acceleration of the car as a function
of time .t
b. Find the velocity and acceleration at sec.t = 1
Soln. We have found earlier (t) t . s′ = 4 + 5
a. The velocity of the car is (instantaneous)rate of change of
position of the car, so .(t) (t) tv = s′ = 4 + 5
The acceleration of the car is the (instantaneous) rate of
change of velocity of the car, so . Note we have not(t) (t)a = v′ = 4
found , it is an exercise for the students.(t)v′
b. In particular velocity of the car at issect = 1
(1) (1) m/secv = 4 + 5 = 9
And the acceleration of the car at sec is .t = 1 (1) m/seca = 4 2