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course MTH 173
10/16 16:30
TEXT SECTION 1.7PROBLEM 1-10
Given function ( e^sinθ)/cosθ for the interval [ - pi/4 , pi/4 ].
Any exponential function is always continuous and the cosθ function is also continuous. Thus is the given function is like f(θ) = e^sinθ and g(θ) = cosθ thus f(θ)/g(θ) is continuous for all θ except when g(θ) is 0. Thus cosθ = 0 , for θ between - pi/4 and pi / 4. Since we can see from the cosθ graph that the function is never 0 for the interval [ -pi/4 , pi/4 ] and thus the given function f(θ) / g(θ) is always continuous.
PROBLEM 1-11
Given f(x) = x^3 + x^2 - 1 for the interval [0,1]
f(0) = 0 + 0 - 1 = -1 and f(1) = 1 + 1 - 1 = 1
Since the function changes it’s sign from x = 0 to x = 1, thus the function will have a x = c such that f(c) = 0, only if the function is continuous for the given interval. A polynomial function with a denominator of 1 is always continuous and thus it has a x = c for which f(c) = 0.
PROBLEM 1-12
Given function f(x) = e^x - 3x for the interval [0,1]
f(0) = e^0 - 0 = 1 - 0 = 1
f(1) = e^1 - 3 = e - 3, the value of e is approx 2.72.
thus f(1) = 2.72 - 3 = -.28 approximately.
Since the function e^x is continuous for all x and 3x is also continuous for all x thus the function f(x) is always continuous. Thus since it changes it’s sign for x values from x = 0 to x = 1, thus the function has a x = c such that f(c) = 0.
PROBLEM 1-15
(a) Given f(x) = x for x < 1 and f(x) = x^2 for 1 < x.
The function x is continuous for all values of x, thus f(x) is continuous for x < 1
Similarly x^2 is continuous for all values of x thus f(x) is continuous for 1 < x
For the entire f(x) to be continuous the value of the function x and x^2 should be same at x = 1
For f(x) = x, f(1) = 1 and for f(x) = x^2, f(1) = 1^2 = 1.
Thus the function f(x) is continuous for all x.
(b) Given g(x) = x for x < 3 and g(x) = x^2 for 3 < x
The function x is continuous for all values of x, thus g(x) is continuous for x < 3
Similarly x^2 is continuous for all values of x thus f(x) is continuous for 3 < x
For the entire f(x) to be continuous the value of the function x and x^2 should be same at x = 3
For f(x) = x, f(3) = 3 and for f(x) = x^2, f(3) = 3^2 = 9.
Since the values of the function are not same, thus the function f(x) is not continuous at x = 3.
PROBLEM 1-16
(a) The quantity of gas in the tank of the car on a journey between New York and Boston continuously decreases, every unit of time since New York has a quantity of gas associated with it till Boston thus the function is continuous.
(b) The number of students enrolled in a class during a semester will have a certain number of students associated with each semester, there is no relation of a value with that of the previous semester. For example if there are 40 students in one semester and 50 the next, the graph for the function will be a point graph, it will not join the 2 points of 40 students and 50 students and thus will not be a continuous graph.
(c) the age of the oldest person alive has a similar interpretation as the number of students enrolled and thus will also not be a continuous function.
PROBLEM 1-20
Given f(x) = kx for 0 < x < 2 and f(x) = 3*(x^2) for 2 < x
The function kx is a straight line is a continuous function and thus is continuous for all value of x thus f(x) is continuous for 0 < x < 2 and so is the case for the function 3*(x^2).
For the function f(x) to be continuous throughout the function kx and 3*(x^2) should have same values at x = 2, which connects the 2 parts of the function f(x).
Thus for f(x) = 3*(x^2) f(2) = 3*2*2 = 12 and for f(x) = kx f(2) = 2k
For the function to be continuous 12 = 2k, thus k = 6.
Thus the value of k = 6 makes the function continuous for all x.
PROBLEM 1-24
Given h(x) = kx for 0 < x < 1 and h(x) = x + 3 for 1 < x < 5
The function kx is a straight line is a continuous function and thus is continuous for all value of x thus h(x) is continuous for 0 < x < 1 and so is the case for the function x + 3.
For the function h(x) to be continuous throughout the function kx and x + 3 should have same values at x = 1, which connects the 2 parts of the function h(x).
Thus for h(x) = kx h(1) = k and for h(x) = x + 3 f(1) = 4
For the function to be continuous 4 = k, thus k = 4.
Thus the value of k = 4 makes the function continuous for all x.
TEXT SECTION 1.8
PROBLEM 1-6
Given function lim x tending to + infinity f(x) = 1 and lim x tending to -infinity f(x) = + infinity
We know the the function e^x tends to + infinity for x tending to + infinity and tends to 0 for x tending to - infinity. If we take the mirror image of the graph about the y axis and shift the function by 1 units in the positive y direction the function f(x) tends to + infinity for x tending to - infinity and the function f(x) tends to 1 for x tending to + infinity. Thus the function is f(x) = e^-x + 1
PROBLEM 1-9
Given function lim x tending to 3 f(x) = 5 and lim x tending to - infinity f(x) = + infinity.
For a normal parabola y = a* x^2, where a is an arbitrary positive constant, the function tends to + infinity when x tends to - infinity. For the same function to tend to 5 when x tends to 3, we need to shift the vertex to the point (3,5).
Thus f(x) = a*(x-3)^2 + 5
PROBLEM 1-12
Given function f(x) = x^5 + 25*x^4 - 37*x^3 - 200*x^2 + 48x + 10.
The function is a 5 degree polynomial which is an odd polynomial
For all odd polynomials as x tends to + infinity f(x) tends to + infinity and as x tends to - infinity f(x) tends to - infinity.
PROBLEM 1-15
Given function f(x) = 25*(e^0.08x)
Every exponential function of the form a*e^kx tends to + infinity as x tends to + infinity and tends to 0 as x tends to - infinity.
PROBLEM 1-18
Given function f(x) = |x| / x
The x tends from the right hand side that is from the positive x side |x| basically tends to x and thus the ratio x/x tends to 1 where when x tends from the left hand side that is form the negative x side |x| basically tends to -x and thus the ration -x/x tends to -1.
Thus the function has a right hand limit of 1 and a left hand limit of -1 and since it is not same thus is not continuous.
PROBLEM 1-20
Given function lim θ tending to 0 ( cos θ - 1 ) / θ
Using the graph to calculate the limit is basically using the values of the function to calculate the limit.
Thus we take θ values in the vicinity of θ = 0.
Let f(θ) = (cos θ - 1 ) / θ
Thus f(0.1) = ( 0.995 - 1 ) / 0.1 = -0.005/0.1 = -0.05
f(0.01) = ( 0.99995 - 1 ) / 0.01 = -0.00005/0.01 = -0.005
f(0.001) = ( 0.999995 - 1) / 0.001 = -0.000005 / 0.001 = -0.0005
f(-0.1) = ( 0.995 - 1 ) / -0.1 = -0.005/-0.1 = 0.05
f(-0.01) = ( 0.99995 - 1 ) / -0.01 = -0.00005/-0.01 = 0.005
f(-0.001) = ( 0.999995 - 1) / -0.001 = -0.000005 / -0.001 = 0.0005
so as we move more and more near θ = 0 from either side on the graph we see that the value of the function tends to 0 thus the limit of the function at θ = 0 is 0.
PROBLEM 1-21
Given lim θ tending to 0 f(θ) = sin θ / θ
Using the graph to calculate the limit is basically using the values of the function to calculate the limit.
Thus we take θ values in the vicinity of θ = 0.
Thus f(0.1) = 0.0998 / 0.1 = 0.998
f(0.01) = 0.009999/0.01 = 0.9999
f(-0.1) = -0.0998/-0.1 = 0.998
f(-0.01) = -0.009999/-0.01 = 0.9999
so as we move more and more near θ = 0 from either side on the graph we see that the value of the function tends to 1 thus the limit of the function at θ = 0 is 1.
PROBLEM 1-24
Given lim h tending to 0 f(h) = (e^5h - 1) / h
Using the graph to calculate the limit is basically using the values of the function to calculate the limit.
Thus we use the values of h in the vicinity of h = 0.
Thus f(0.1) = (e^.5 - 1)/.1 = (1.6487 - 1)/.1 = 0.6487/.1 = 6.487
f(0.01) = (e^.05 - 1)/0.01 = (1.05127 - 1)/.01 = 5.127
f(0.001) = (e^0.005 - 1)/0.001 = (1.00501 - 1)/0.001 = 5.01
f(-0.1) = (e^-.5 - 1)/-.1 = (0.60653 - 1)/-.1 = 0.39347/.1 = 3.9347
f(-0.01) = (e^-.05 - 1)/-.01 = (0.95123 - 1)/-.01 = 0.04877/.01 = 4.877
f(-0.001) = (e^-.005 - 1)/-.001 = (0.995012 - 1)/-.001 = 0.004988/0.001 = 4.988
thus as we see that as h tends to 0 from the both the sides, that is the right hand and the left hand side the function tends to the value 5 that is the limit of the function f(h) as h tends to 0 is 5.
PROBLEM 1-29
Given function f(x) = | x - 4 | / ( x - 4 ) and a = 4.
Thus the right hand limit that is lim x tending to 4 + f(x) = + ( x - 4 ) / (x - 4 ) = 1 and
The left hand limit that is lim x tending to 4 - f(x) = - ( x - 4 ) / ( x - 4 ) = -1
Since the left hand limit and the right hand limit is different thus the limit of the function at x = 4 does not exists.
The graph for the same function gives no value of the limit. That is for x greater than 4 the value of the function is 1, thus as the value of x tends from the right hand side the value of the function tends to 1 and for x less than 4 the value of the function is -1, thus as we tend the graph from the left hand side the value of the function tends from -1. Thus there is no defined value of the function at x = 4 for the function, thus no limit of the function at x = 4.
PROBLEM 1-32
Given function f(θ) = sinθ/θ
For it to be stay within 0.001 of 1, we set the range of the function to lie with 0.999 and 1.001
The value of the function f(θ) is always within the 1.001. Using the graph and the calculator that the value of the function at θ = 0.07 is 0.99918 which is within the y value range of 0.999 and 1.001, and on the same side the value of the function at θ = -0.07 is 0.99918. Thus the estimate the value of θ is 0.07 around θ = 0 on both the sides, for the value of the function to be within 0.001 of 1.
PROBLEM 1-35
Given function f(x) = |x|/x for x ≠ 0 and f(x) = 0 for x = 0.
The function for all x > 0 is 1 and thus is continuous for x > 0
For all x<0 the function value is -1 and thus is continuous for x < 0.
For x = 0 the value of the function is 0.
For lim x tending to + 0 the value of the function is 1 and for lim x tending to - 0 the value of the function is -1. Since the value of the right hand limit, left hand limit and the actual value does not match thus the function is discontinuous at x = 0.
PROBLEM 1-37
Given f(x) = 2x/x for x ≠ 0 and f(x) = 3 for x = 0.
The function f(x) = 2x/x = 2 for all x except x = 0. Thus lim x tending to 0 f(x) = 2 but the value of the function at x = 0, f(0) = 3. Thus the function is discontinuous at x = 0, thus the function is continuous for all x except x = 0.
PROBLEM 1-42
To calculate lim x tending to 0+ x^x.
f(x) = x ^ x
We calculate the limit by using smaller and smaller positive values of x.
Thus f(0.1) = 0.1^0.1 = 0.79432
f(0.01) = 0.01 ^ 0.01 = 0.954992
f(0.001) = .001 ^ .001 = 0.993116
thus as we see that as we get closer and closer to x = 0, the value keeps tending towards 1.
Thus lim x tending to 0+ x^x = 1.
PROBLEM 1-45
Given lim x tending to infinity f(x) = L
According to the definition of limit we define it as a number L( if one exists ) such that for every epsilon > 0(as small as we want), there is a delta > 0 (sufficiently small) such that if |x - infinity| < delta and x is not infinity, then | f(x) - L | < epsilon.
Another way to define the function is by substituting x by 1/h. thus we have f(h) = f(1/x) and lim x tending to infinity = lim h tending to 0, the L remains the same L.
Thus we have lim h tending to 0 f(h) = L
Thus we would define the limit as a number L( if one exists ) such that for every epsilon > 0(as small as we want), there is a delta > 0 (sufficiently small) such that if |h - 0| < delta and x is not infinity, then | f(h) - L | < epsilon where h = 1/x.
PROBLEM 1-46
(a)
Given f(x) = 3x + 1
When x = 0.1 , f(0.1) = 3(.1) + 1 = 1.3
x = 0.01 f(0.01) = 3(0.01) + 1 = 1.03
x = 0.001 f(0.001) = 3(0.001) + 1 = 1.003
x = 0.0001 f(0.0001) = 3(0.0001) + 1 = 1.0003
x = -0.1 f(-0.1) = 3(-0.1) + 1 = 0.7
x = -0.01 f(-0.01) = 3(-0.01) + 1 = 0.97
x = -0.001 f(-0.001) = 3(-0.001) + 1 = 0.997
x = -0.0001 f(-0.0001) = 3(-0.0001) + 1 = 0.9997
(b)
As the value of x gets closer to x = 0 from the right hand side the value of the function tends to 1 and as value of the x gets closer to x = 0 from the left hand side the value of the function tends to 1.
Thus the value of the function f(x) as lim x tends to 0 = 1.
(c)
The graph of the function is f(x) = 3x + 1.
The given function is a straight line, the slope of which is 3 and has a y intercept of 1. The value of the function at x = 0 is 1. Thus from the graph we see that as we approach the graph from the right hand side that the value that we approach is 1 and again as we approach the value of the function from left hand side that is for x less than 0 the function approaches 1.
(d)
The value of conjectured limit is 1 and the difference between the value of the function and the limit is to be 0.01. Thus we have the y value range of 0.99 to 1.01. The function f(x) = 3x + 1. Thus the value of x for the corresponding y range will be
0.99 = 3x + 1
-0.01 = 3x, x = - 0.0034
And 1.01 = 3x + 1
0.01 = 3x, x = 0.01/3 = 0.0034
Thus the interval of x near 0 for which the y value of the function is less than 0.01 from the conjectured limit value is 0.0034 near 0.
PROBLEM 1-49
(a)
Given function f(x) = sin3x
x = 0.1, f(0.1) = sin(0.3) = 0.295
x = 0.01, f(0.01) = sin(0.03) = 0.029995
x = 0.001, f(0.001) = sin(0.003) = 0.0029999
x = 0.0001, f(0.0001) = sin(0.0003) = 0.000299999
x = -0.1, f(-0.1) = sin(-0.3) = -0.295
x = -0.01, f(-0.01) = sin(-0.03) = -0.029995
x = -0.001, f(-0.001) = sin(-0.003) = -0.0029999
x = -0.0001, f(-0.0001) = sin(-0.0003) = -0.000299999
(b)
As we see that as we tend x values closer to 0 from both right and the left side the value of the function tends to 0 thus the limit of the function is 0 as x tends to 0.
(c)
The graph of this function is a typical sinx function with the only difference that the period of the function is 2pi/3 instead of 2pi. The value of the function sin3x is 0 at x = 0. The value of the function tends to 0 as x tends from the right hand side that is from positive values of x and the value of the function tends to 0 as x tends form the left hand side that is from the negative values of x as seen from the graph, thus the graph too confirms the same limit.
(d)
The value of conjectured limit is 0 and the difference between the value of the function and the limit is to be 0.01. Thus we have the y value range of -0.01 to 0.01. The function f(x) = sin(3x).
Thus for y value 0.01, sin(3x) = 0.01, 3x = 0.01, x = 0.0034
For y value -0.01, sin(3x) = -0.01, 3x = -0.01, x = -0.0034
Thus for x value 0.0034 units on either side of x = 0 gives us a range for which the function has a value such that its difference from the conjectured limit is less than 0.01.
PROBLEM 1-52
(a)
Given function f(x) = (e^x - 1) / x
x = 0.1, f(0.1) = (e^0.1 - 1 ) / 0.1 = 0.1051/0.1 = 1.051
x = 0.01, f(0.01) = (e^0.01 -1)/0.01 = 0.01005/0.01 = 1.005
x = 0.001, f(0.001) = (e^0.001 - 1)/0.001 = 0.0010005/0.001 = 1.0005
x = 0.0001, f(0.0001) = (e^0.0001 - 1)/0.0001 = 0.000100005/0.0001 = 1.0005
x = -0.1, f(-0.1) = (e^-0.1 - 1)/-0.1 = 0.095162/0.1 = 0.95162
x = -0.01, f(-0.01) = (e^-0.01 - 1)/-0.01 = 0.0099501/0.01 = 0.99501
x = -0.001, f(-0.001) = (e^-0.001 - 1)/-0.001 = 0.00099950/0.001 = 0.99950
x = -0.0001, f(-0.0001) = (e^-0.0001 - 1)/-0.0001 = 0.000099995/0.0001 = 0.99995
(b)
As x tends to the value of x = 0 from the right hand side that is the positive values of x and the left hand side that is the negative value of x we see that the value of the function tends towards 1. Thus the limit of the function as x tends to 0 is 1.
(c)
The graph of the given function is almost like an e ^ x function when zoomed in and seen. Thus for the given function we see that as we approach the vertical line x = 0 from the positive values of of x we approach the value of the function as 1, similarly as we approach the same x = 0 line form the left hand side that is from the negative values of x we again approach the value of the function as 1.
(d)
The value of conjectured limit is 1 and the difference between the value of the function and the limit is to be 0.01. Thus we have the y value range of 0.99 to 1.01. The function is f(x) = (e^x - 1)/x. Using the graph and calculator we obtain that the value of the function at x = 0.01 is 1.005 which is within the y range of 0.99 and 1.01, also the value of the function at x = -0.01 is 0.995 which lies within the y value of 0.99 and 1.01. Thus for an interval of 0.01 within 0 on both the side of 0, the difference between the conjectured limit and the value of the function is less than 0.01.
PROBLEM 1-54
Given function f(x) = ( x + 3 ) / ( 2 - x )
To find limit of the function for x tending to infinity, we replace x by 1 / x,
Thus the function become ( 1 + 3x) / ( 2x - 1) with lim x tending to 0.
Thus the value of the limit using the limit rules will be
lim x tending to 0 ( 1 + 3x) / lim x tending to 0 ( 2x - 1 ) = 1 / -1 = -1.
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