Exercise 1
Study the following functions and determine if they are continuous. If not, state where the discontinuities exist and what type they are:
1 
2 
3 
4 
5 
6 
Exercise 2
Determine if the following function is continuous at x = 0.
Exercise 3
Determine if the following function is continuous on (0,3). If not, state where the discontinuities exist and what type they are:
Exercise 4
Are the following functions continuous at x = 0?
Exercise 5
Given the function:
1 Prove that f(x) is not continuous at x = 5.
2Is there a continuous function which coincides with f(x) for all values with the exception x = 5? If so, determine the function.
Exercise 6
Determine if the following function is continuous. If not, state where the discontinuities exist or why the function is not continuous:
Exercise 7
Determine if the following function is continuous at x = 0.
Exercise 8
Determine the value of a to make the following function continuous.
Exercise 9
The function defined by:
is continuous on [0, ∞).
Determine the value of a that would make this statement true.
Solution of exercise 1
Study the following functions and determine if they are continuous. If not, state where the discontinuities exist:
1
The function is continuous at all points of its domain.
D = R − {−2,2}
The function has two points of discontinuity at x = −2 and x = 2.
2
The function is continuous at R with the exception of the values that annul the denominator. If this is equal to zero and the equation is solved, the discontinuity points will be obtained.
x = −3; and by solving the quadratic equation: 
and 
are also obtained
The function has three points of discontinuity at 
, and .
3
\lim _ {x \rightarrow 2^{-}} (x + 1) = 3
\lim _ {x \rightarrow 2^{+}} (2x - 1) = 3
The function is continuous.
4
\lim _ {x \rightarrow 0^{-}} (x^2 - 1) = -1
\lim _ {x \rightarrow 0^{+}} (2x - 3) = -3
The function has a jump discontinuity at x = 0 .
5
\lim_{x \rightarrow 1^{-}} (\frac{1}{x}) = 1
\lim_{x \rightarrow 1^{+}} \sqrt{x + 1} = \sqrt{2}
The function has a jump discontinuity at x = 1 .
6
\lim_{x \rightarrow 0^{-}} (\frac {e^x}{e^x + 1}) = \frac{1}{2}
\lim_{x \rightarrow 0^{+}} x^2 + 1 = 1
The function has a jump discontinuity at x = 1/2 .
Solution of exercise 2
Determine if the following function is continuous at x = 0.
\lim_{x \rightarrow 0^{+}} x^ {\frac{1}{x}} = 0 ^ {\frac{1}{0^{+}}} = 0 ^{\infty} = 0
\lim_{x \rightarrow 0^{-}} x^ {\frac{1}{x}}
At x = 0, there is an essential discontinuity.
Solution of exercise 3
Determine if the following function is continuous on (0,3). If not, state where the discontinuities exist and what type they are:
\lim_{x \rightarrow 1^{-}} x^2 = 1
\lim_{x \rightarrow 1^{+}} 0 = 0
At x = 1, there is a jump discontinuity.
\lim_{x \rightarrow 2^{-}} 0 = 0
\lim_{x \rightarrow 2^{+}} x - 1 = 1
At x = 2, there is a jump discontinuity.
Solution of exercise 4
Are the following functions continuous at x = 0?
\lim_{x \rightarrow 0^{-}} 2^{-x} = 2 ^{0^{-}} = 2^0 = 1
\lim_{x \rightarrow 0^{+}} 2^{-x} = 2^{-0^{+}} = \frac{1}{2^0} = 1
The function is continuous at x = 0.
Solution of exercise 5
Given the function:
1 Prove that f(x) is not continuous at x = 5.
\lim_{x \rightarrow 5} \frac{x^2 - 25}{x - 5} = \frac{0}{0}
Solve the indeterminate form.
\lim_{x \rightarrow 5} \frac{(x + 5) (x - 5)}{x - 5} = \lim_{x \rightarrow 5} (x + 5) = 10
f (x) is not continuous at x = 5 because:
\lim_{x \rightarrow 5} f(x) \neq f(5)
2 Is there a continuous function which coincides with f(x) for all values with the exception x = 5? If so, determine the function.
If \lim_{x \rightarrow 5} f(x) = f(5) = 10 the function would be continuous, then the function is redefined:
Solution of exercise 6
Determine if the following function is continuous. If not, state where the discontinuities exist or why the function is not continuous:
The function f(x) is continuous for x ≠ 0. Therefore, study the continuity at x = 0.
\lim_{x \rightarrow 0 ^{-}} \frac{x + 1}{|x|} = \lim_{x \rightarrow 0^{-}} \frac{x + 1}{-x} = \lim_{x \rightarrow 0^{-}}(-1 - \frac{1}{x}) = -1 - \frac{1}{0^{-}} = \infty
\lim_{x \rightarrow 0 ^{+}} \frac{x + 1}{|x|} = \lim_{x \rightarrow 0^{+}} \frac{x + 1}{x} = \lim_{x \rightarrow 0^{+}}(-1 - \frac{1}{x}) = -1 - \frac{1}{0^{+}} = \infty
The function is not continuous at x = 0, because it is defined at that point.
Solution of exercise 7
Determine if the following function is continuous at x = 0:
The function 
is bounded by 
, 
, therefore takes place:
\lim_{x \rightarrow{0}} (x \cdot sin \dfrac{1}{x}) = 0, since any number multiplied by zero gives zero.
As f(0) = 0.
The function is continuous.
Solution of exercise 8
Determine the value of a to make the following function continuous:
\lim_{x \rightarrow 1^{-}} (x + 1) = 2
\lim_{x \rightarrow 1^{+}} (3 - ax^2) = 3 - a
Solution of exercise 9
The function defined by:
is continuous on [0, ∞).
Determine the value of a that would make this statement true.
\lim_{x \rightarrow 8^{-}} \sqrt{ax} = \sqrt{8a}
\lim_{x \rightarrow 8^ {+}} \frac{x^2 - 32}{x - 4} = 8
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