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Solved Problems on Limits and Continuity. Overview of Problems. 2. 1. 4. 3. 5. 6. 7. 8. 9. 10. Overview of Problems. 11. 12. 13. 14. 15. Main Methods of Limit Computations. The following undefined quantities cause problems:. 1. In the evaluation of expressions, use the rules. - PowerPoint PPT Presentation
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Solved Problems on Limits and Continuity
Calculators
Overview of Problems
2
0
sinlim
sinx
xx x
12
2
3 2lim2x
x xx
23 2
3 2
1lim3 5 2x
x x xx x x
3 2 2lim 1 1x
x x
2 2lim 1 1x
x x x x
2 20
2lim2 1 3 1x
xx x x x
0
sin 3lim
6x
xx
0
sin sinlimx
xx
4
5 6
7 8
9 10
2 20
2sinlim
2sin 1 sin 1x
x x
x x x x tan x
2
lim ex
Calculators
Overview of Problems11 Where tan is continuous?y x
12
2
1Where f sin is continuous?1
13
2
How must f 0 be determined so that f , 0, 1
is continuous at 0?
x xx xx
x
14
2
0 0
0
Which of the following functions have removablesingularities at the indicated points?
2 8 1a) f , 2, b) g , 12 1
1c) h sin , 0
x x xx x x xx x
t t tt
Show that the equation sin e has many solutions.xx15
Calculators
Main Methods of Limit Computations
If the function, for which the limit needs to be computed, is defined by an algebraic expression, which takes a finite value at the limit point, then this finite value is the limit value.
3
If the function, for which the limit needs to be computed, cannot be evaluated at the limit point (i.e. the value is an undefined expression like in (1)), then find a rewriting of the function to a form which can be evaluated at the limit point.
4
In the evaluation of expressions, use the rules 2
0, , negativenumber .positive numbera
The following undefined quantities cause problems: 10 000 , , , , 0 , .0
Calculators
Main Computation Methods
If a square root appears in the expression, then multiply and divide by the conjugate of the square root expression.
3
1 2 1 21 2
1 21 2 3 01 2 1 2 x
x x x xx x
x xx xx x x x
Cancel out common factors of rational functions.2 2
11 11 1 2.1 1 x
x xx xx x
Use the fact that4 0
sinlim 1.x
xx
Frequently needed rule1 2 2.a b a b a b
Calculators
Continuity of FunctionsFunctions defined by algebraic or elementary expressions involving polynomials, rational functions, trigonometric functions, exponential functions or their inverses are continuous at points where they take a finite well defined value.
1
If f is continuous, f(a) < 0 and f(b) > 0, then there is a point c between a and b so that f(c) = 0.
4
A function f is continuous at a point x = a if2 limff .
x ax a
The following are not continuous x = 0:3 1 1f , g sin ,h .xx x x
x x x
Intermediate Value Theorem for Continuous Functions
Used to show that equations have solutions.
Calculators
Limits by RewritingProblem 1
2
2
3 2lim2x
x xx
Solution 2 1 23 2Rewrite 1.2 2
x xx x xx x
2
2 2
3 2Hence lim lim 1 1.2x x
x x xx
Calculators
Limits by RewritingProblem 2
3 2
3 2
1lim3 5 2x
x x xx x x
Solution
3 2 2 3
3 2
2 3
1 1 111 1.3 5 23 5 2 1
x
x x x x x xx x x
x x x
Calculators
Limits by RewritingProblem 3 2 2lim 1 1
xx x
Solution
2 2 2 2
2 2
2 2
1 1 1 11 1
1 1
x x x xx x
x x
2 2
2 2 2 2
2 2 2 2 2 2
1 1 1 1 2
1 1 1 1 1 1
x x x x
x x x x x x
Rewrite
2 2
2 2
2Hence lim 1 1 lim 0.1 1x x
x xx x
Calculators
Limits by RewritingProblem 4 2 2lim 1 1
xx x x x
Solution
2 2
2 2 2 2
1 1 2 2
1 1 1 1
x x x x xx x x x x x x x
2 2
2 22 2
2 2
1 1
1 1 1 11 1
x x x x
x x x xx x x xx x x x
2 2
222
1 1 1 11 1x
x
x x x x
Rewrite
Next divide by x.
Calculators
Limits by RewritingProblem 5 2 20
2lim2 1 3 1x
xx x x x
Solution
2 2 2 2
2 2 22 2
2 2 1 3 1 2 2 1 3 1
42 1 3 1
x x x x x x x x x x
x xx x x x
2 2
2 2
2 2 2 2
2
2 1 3 1
2 2 1 3 1
2 1 3 1 2 1 3 1
xx x x x
x x x x x
x x x x x x x x
2 2
0
2 2 1 3 11
4 x
x x x x
x
Rewrite
Next divide by x.
Calculators
Limits by RewritingProblem 6
0
sin 3lim
6x
xx
Solution
sin 3 sin 31Rewrite 6 2 3x xx x
0
sinUse the fact that lim 1.
0 0
sin 3 sin 3 1Since lim 1, we conclude that lim .3 6 2x x
x xx x
Calculators
Limits by RewritingProblem 7
0
sin sinlimx
xx
Solution
0
0
sinsince lim 1. In the above, that fact
was applied first by substituting sin .
sin sinHence lim 1.
sinx
x
xx
0
sin sin sin sin sin1
sin x
x x xx x x
Rewrite:
Calculators
Limits by RewritingProblem 8
2
0
sinlim
sinx
xx x
Solution
2 2
02
sin sin1
sin sin x
x x xx x x x
Rewrite:
Calculators
Limits by RewritingProblem 9
2 20
2sinlim
2sin 1 sin 1x
x x
x x x xSolution
2 2
2 2
2 2 2 2
2sin
2sin 1 sin 1
2sin 2sin 1 sin 1
2sin 1 sin 1 2sin 1 sin 1
x x
x x x x
x x x x x x
x x x x x x x x
Rewrite
2 2
2 2
2 2
2 2
2sin 2sin 1 sin 1
2sin 1 sin 1
2sin 2sin 1 sin 1
sin 2sin
x x x x x x
x x x x
x x x x x x
x x x x
Calculators
Limits by RewritingProblem 9
2 20
2sinlim
2sin 1 sin 1x
x x
x x x xSolution(cont’d)
2 2
2 2
2 2
2sin
2sin 1 sin 1
2sin 2sin 1 sin 1
sin 2sin
x x
x x x x
x x x x x x
x x x x
Rewrite
2 2sin1 2 2sin 1 sin 1
sin sinsin 2 1
xx x x x
xx x
x xx x
0
3 2 2.2 1x
Here we used the fact that all sin(x)/x terms approach 1 as x 0.
Next divide by x.
Calculators
One-sided Limits Problem 10 tan x
2
lim ex
Solution
2
tan
2
For , tan 0 and lim tan .2
Hence lim e 0.
x
x
x
x x x
Calculators
Continuity Problem 11 Where the function tan is continuous?y x
Solution
sinThe function tan is continuous whenever cos 0.
cos
Hence tan is continuous at , .2
xy x x
x
y x x n n
Calculators
Continuity Problem 12 2
1Where the function f sin is continuous?1
Solution 2
1The function f sin is continuous at all points1
where it takes finite values.
2 2
1 1If 1, is not finite, and sin is undefined.1 1
2 2
1 1If 1, is finite, and sin is defined and also finite.1 1
2
1Hence sin is continuous for 1.1
Calculators
Continuity Problem 13
2
How must f 0 be determined so that the function
f , 0, is continuous at 0?1
x xx x xx
Solution
0
0
0 0
2
0 0 0
Condition for continuity of a function f at a point is:
limf f . Hence f 0 must satisfy f 0 lim f .
1Hence f 0 lim lim lim 0.
1 1
x x x
x x x
xx x x
x xx x xx x
Calculators
Continuity
0
0
A number for which an expression f either is undefined or
infinite is called a of the function f . The singularity is said to be , if f can be defi
singularityremovab ned in such a way le that
x x
x
0
the function f becomes continous at .x x
Problem 14
2
0
0
0
Which of the following functions have removablesingularities at the indicated points?
2 8a) f , 22
1b) g , 11
1c) h sin , 0
x xx xx
xx xx
t t tt
Answer
Removable
Removable
Not removable
Calculators
Continuity Show that the equation sin e has
inifinitely many solutions.
xx
Observe that 0 e 1 for 0, and that sin 1 , .2
nx x n n
Hence f 0 for if is an odd negative number 2
and f 0 for if is an even negative number.2
x x n n
x x n n
Problem 15
Solution sin e f sin e 0.x xx x x
By the intermediate Value Theorem, a continuous function takes any value between any two of its values. I.e. it suffices to show that the function f changes its sign infinitely often.
We conclude that every interval 2 , 2 1 , and 0, contains 2 2
a solution of the original equation. Hence there are infinitely many solutions.
n n n n