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 f(x) = \frac{5}{x^4 - 16}

2 f(x) = \frac{x - 7}{x^3 - x^2 - 11x + 3}

3 f(x) = \left\{\begin{matrix} x^2 - 1  if x \leq 0 \\ 2x - 3, if x>0 \end{matrix}\right

4 f(x) = \left\{\begin{matrix} x^2 - 1  if x \leq 0 \\2x - 3, if x>0 \end{matrix}\right

5 f(x) = \left\{\begin{matrix} \frac{1}{x}  if x < 1 \\\sqrt{x + 1}, if x\geq 1 \end{matrix}\right

6 f(x) = \left\{\begin{matrix} \frac{e^x}{e^x + 1}  if x \leq 0 \\x^2 + 1, if x> 1 \end{matrix}\right

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Exercise 2

Determine if the following function is continuous at x = 0.

f(x) = \left\{\begin{matrix} x ^ {\frac{1}{x}},  if x > 0 \\0, if x = 0 \end{matrix}\right

Exercise 3

Determine if the following function is continuous on (0,3). If not, state where the discontinuities exist and what type they are:

f(x) = \left\{\begin{matrix} x^2,  if0 < x < 1 \\0, if 1 \leq x < 2, x - 1, if 2 \leq x < 3 \end{matrix}\right

Exercise 4

Are the following functions continuous at x = 0?

f(x) = 2 ^ {-x}

Exercise 5

Given the function:

f(x) = \left\{\begin{matrix} \frac{x^2 - 25}{x - 5},  if x \neq 5 \\0, if x= 5\end{matrix}\right

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:
f(x) = \frac{x + 1}{|x|}

 

Exercise 7

Determine if the following function is continuous at x = 0.

f(x) = \left\{\begin{matrix} x \cdot sin \frac{1}{x}  if x \neq 0 \\0, if x = 0 \end{matrix}\right

Exercise 8

Determine the value of a to make the following function continuous.

f(x) = \left\{\begin{matrix} x + 1, if x \leq 1 \\3 - ax^2, if x> 1 \end{matrix}\right

Exercise 9

The function defined by:

f(x) = \left\{\begin{matrix} \sqrt{ax}  if 0 \leq x \leq 8 \\\frac{x^2 - 32}{x - 4}, if x> 8 \end{matrix}\right

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 f(x) = \frac{5}{x^4 - 16}

The function is continuous at all points of its domain.

D = R - {-2, 2}

D = R − {−2,2}

The function has two points of discontinuity at x = −2 and x = 2.

 

2  f(x) = \frac{x - 7}{x^3 - x^2 - 11x + 3}

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 - x^2 - 11x + 3 = 0

x = −3; and by solving the quadratic equation: x = 2 - \sqrt{3} and x = 2 + \sqrt{3} are also obtained

The function has three points of discontinuity at x = 3, x = 2 - \sqrt{3} and x = 2 + \sqrt{3}.

 

3  f(x) = \left\{\begin{matrix} x^2 - 1  if x \leq 0 \\ 2x - 3, if x>0 \end{matrix}\right

f(2) = 3

    \[\lim _ {x \rightarrow 2^{-}} (x + 1) = 3\]

    \[\lim _ {x \rightarrow 2^{+}} (2x - 1) = 3\]

The function is continuous.

4  f(x) = \left\{\begin{matrix} x^2 - 1  if x \leq 0 \\2x - 3, if x>0 \end{matrix}\right

f(0) = -1

    \[\lim _ {x \rightarrow 0^{-}} (x^2 - 1) = -1\]

    \[\lim _ {x \rightarrow 0^{+}} (2x - 3) = -3\]

The function has a jump discontinuity at x = 0 .

 

 

5 f(x) = \left\{\begin{matrix} \frac{1}{x}  if x < 1 \\\sqrt{x + 1}, if x\geq 1 \end{matrix}\right

f(1) = \sqrt{2}

    \[\lim_{x \rightarrow 1^{-}} (\frac{1}{x}) = 1\]

    \[\lim_{x \rightarrow 1^{+}} \sqrt{x + 1} = \sqrt{2}\]

The function has a jump discontinuity at x = 1 .

 

f(x) = \left\{\begin{matrix} \frac{e^x}{e^x + 1}  if x \leq 0 \\x^2 + 1, if x> 1 \end{matrix}\right

f(0) = \frac{1}{2}

    \[\lim_{x \rightarrow 0^{-}} (\frac {e^x}{e^x + 1}) = \frac{1}{2}\]

    \[\lim_{x \rightarrow 0^{+}} x^2 + 1 = 1\]

The function has a jump discontinuity at x = 1/2 .

 

Solution of exercise 2

Determine if the following function is continuous at x = 0.

f(x) = \left\{\begin{matrix} x ^ {\frac{1}{x}},  if x > 0 \\0, if x = 0 \end{matrix}\right

f(0) = 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:

f(x) = \left\{\begin{matrix} x^2,  if0 < x < 1 \\0, if 1 \leq x < 2, x - 1, if 2 \leq x < 3 \end{matrix}\right

f(1) = 0

    \[\lim_{x \rightarrow 1^{-}} x^2 = 1\]

    \[\lim_{x \rightarrow 1^{+}} 0 = 0\]

At x = 1, there is a jump discontinuity.

f(2) = 1

    \[\lim_{x \rightarrow 2^{-}} 0 = 0\]

    \[\lim_{x \rightarrow 2^{+}} x - 1 = 1\]

At x = 2, there is a jump discontinuity.

 

Solution of exercise 4

Are the following functions continuous at x = 0?

f(x) = 2 ^ {-x}

f(0) = 2 ^{-0} = \frac{1}{2^{0}} = 1

    \[\lim_{x \rightarrow 0^{-}} 2^{-x} = 2 ^{0^{-}} = 2^0 = 1\]

    \[\lim_{x \rightarrow 0^{+}} 2^{-x} = 2^{-0^{+}} = \frac{1}{2^0} = 1\]

The function is continuous at x = 0.

 

Solution of exercise 5

Given the function:

f(x) = \left\{\begin{matrix} \frac{x^2 - 25}{x - 5},  if x \neq 5 \\0, if x= 5\end{matrix}\right

1  Prove that f(x) is not continuous at x = 5.

f(5) = 0

    \[\lim_{x \rightarrow 5} \frac{x^2 - 25}{x - 5} = \frac{0}{0}\]

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:

f(x) = \left\{\begin{matrix} \frac{x^2 - 25}{x - 5},  if x \neq 5 \\0, if x= 5\end{matrix}\right

 

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:

f(x) = \frac{x + 1}{|x|}

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:

f(x) = \left\{\begin{matrix} x \cdot sin \frac{1}{x}  if x \neq 0 \\0, if x = 0 \end{matrix}\right

The function sin \frac{1}{x}  is bounded by |sin \frac{1}{x}| \leq 1, x \neq 0,  therefore takes place:

    \[\lim_{x \rightarrow{0}} (x \cdot sin \frac{1}{x}) = 0\]

, 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:

f(x) = \left\{\begin{matrix} x + 1, if x \leq 1 \\3 - ax^2, if x> 1 \end{matrix}\right

f(1) = 2

    \[\lim_{x \rightarrow 1^{-}} (x + 1) = 2\]

    \[\lim_{x \rightarrow 1^{+}} (3 - ax^2) = 3 - a\]

3 - a = 2

a = 1

 

Solution of exercise 9

The function defined by:

f(x) = \left\{\begin{matrix} \sqrt{ax}  if 0 \leq x \leq 8 \\\frac{x^2 - 32}{x - 4}, if x> 8 \end{matrix}\right

is continuous on [0, ∞).

Determine the value of a that would make this statement true.

    \[\lim_{x \rightarrow 8^{-}} \sqrt{ax} = \sqrt{8a}\]

    \[\lim_{x \rightarrow 8^ {+}} \frac{x^2 - 32}{x - 4} = 8\]

\sqrt{8a} = 8

a = 8

 

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.