MHB Finding the value of the constant that makes the function continuous?

[Umar]

Hello, I am finding this questions quite difficult, can someone please offer some insight as to what needs to be done.

View attachment 5964

Do we need to do limit tests to the left and right of x = 7?

 

[soroban]

For what values of C is the following function continuous at x = -7\,?

. . f(x) \;=\; \begin{Bmatrix} \dfrac{\frac{1}{x} + \frac{1}{7}}{x+7} && \text{if }x \ne -7 \\ C && \text{if }x = -7 \end{Bmatrix}

\text{We have: }\;f(x) \;=\; \dfrac{\frac{1}{x} + \frac{1}{7}}{x+7} \;=\;\dfrac{7+x}{7x(x+7)} \;=\; \frac{1}{7x}

\text{Then: }\;f(-7) \;=\;-\frac{1}{7(-7)}

\text{Therefore: }\:C = -\frac{1}{49}

 

[HallsofIvy]

No! You need to look at the right and left limits at x= -7!
(The "no" was in response to Umar's original question, not to Soroban's response.)

 

[HallsofIvy]

With some comments:

\text{We have: }\;f(x) \;=\; \dfrac{\frac{1}{x} + \frac{1}{7}}{x+7} \;=\;\dfrac{7+x}{7x(x+7)} \;=\; \frac{1}{7x}

For all x except x= -7! So the limit, as x goes to -7, is \frac{1}{7(-7)}

In order that this function be continuous we must have the value at -7 the same as that limit:

\text{Then: }\;f(-7) \;=\;\frac{1}{7(-7)}

\text{Therefore: }\:C = -\frac{1}{49}

 

[Umar]

Thank you so much to the both of you, I understand it now.