1. Solved Problems on Limits and Continuity

2. Mika Seppälä: Limits and Continuity
Overview of Problems 1 2 4 3 5 6 7 8 9 10
Mika Seppälä: Limits and Continuity

3. Mika Seppälä: Limits and Continuity
Overview of Problems 11 12 13 14 15
Mika Seppälä: Limits and Continuity

4. Main Methods of Limit Computations1
The following undefined quantities cause problems: 2
In the evaluation of expressions, use the rules
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
Mika Seppälä: Limits and Continuity

5. Main Computation Methods
Frequently needed rule 1
Cancel out common factors of rational functions. 2
If a square root appears in the expression, then multiply and divide by the conjugate of the square root expression. 3
Use the fact that 4
Mika Seppälä: Limits and Continuity

6. Continuity of Functions
Functions 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
A function f is continuous at a point x = a if 2
Used to show that equations have solutions.
The following are not continuous x = 0: 3 4
Intermediate Value Theorem for Continuous Functions
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.
Mika Seppälä: Limits and Continuity

7. Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 1
Solution
Mika Seppälä: Limits and Continuity

8. Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 2
Solution
Mika Seppälä: Limits and Continuity

9. Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 3
Solution
Rewrite
Mika Seppälä: Limits and Continuity

10. Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 4
Solution
Rewrite
Next divide by x.
Mika Seppälä: Limits and Continuity

11. Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 5
Solution
Rewrite
Next divide by x.
Mika Seppälä: Limits and Continuity

12. Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 6
Solution
Mika Seppälä: Limits and Continuity

13. Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 7
Solution
Rewrite:
Mika Seppälä: Limits and Continuity

14. Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 8
Solution
Rewrite:
Mika Seppälä: Limits and Continuity

15. Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 9
Solution
Rewrite
Mika Seppälä: Limits and Continuity

16. Mika Seppälä: Limits and Continuity
Limits by Rewriting
Problem 9
Solution (cont’d)
Rewrite
Next divide by x.
Here we used the fact that all sin(x)/x terms approach 1 as x  0.
Mika Seppälä: Limits and Continuity

17. Mika Seppälä: Limits and Continuity
One-sided Limits
Problem 10
Solution
Mika Seppälä: Limits and Continuity

18. Mika Seppälä: Limits and Continuity
Problem 11
Solution
Mika Seppälä: Limits and Continuity

19. Mika Seppälä: Limits and Continuity
Problem 12
Solution
Mika Seppälä: Limits and Continuity

20. Mika Seppälä: Limits and Continuity
Problem 13
Solution
Mika Seppälä: Limits and Continuity

21. Mika Seppälä: Limits and Continuity
Problem 14
Answer
Removable
Not removable
Mika Seppälä: Limits and Continuity

22. Mika Seppälä: Limits and Continuity
Problem 15
Solution
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.
Mika Seppälä: Limits and Continuity