Section 2.5: Continuity

1. Check if the following function is continuous at the given number  \( x = a \).

(a) \(\displaystyle{ f(x)=\begin{cases} \frac{2x-10}{x-5}, \:\: x \ne 5 \\ 3, \: \: x =5 \end{cases}}\)    at  \( a=5\)   solution

 

(b) \(\displaystyle{ f(x)=\begin{cases} x+1,  \:\:  x \le 2 \\ 2x-1,  \:\:  x >2 \end{cases}}\)    at  \( a=2\)    solution

 

2. Consider the following graph of \(f\). Is \( f \) continuous at \( x = -2, 0, 2, 3\)? If you say Yes, justify your answer using the definition of continuity. If you say No, state what type of discontinuity (removable, jump or infinite) occurs at that value of \(x\).

graph

solution  

 

3. Explain why the function \(f\) is discontinuous at \( 1 \).

\( f(x)=\begin{cases} \dfrac{2x-2}{x^2-1},  \:\: x \ne 1 \\ 2,  \:\: x = 1. \end{cases}\)

solution

 

4. Determine whether the following function is continuous at \( x = a \). If f is discontinuous at \( x = a \), state what type of discontinuity occurs at that value of \(x\).  solution

(a) \(  f(x) = \dfrac{1}{x-3}; \; a =3 \)    (b) \(f(x) = \dfrac{|x|}{x} ; \; a = 0 \)     (c) \( f(x) = \dfrac{x^2 - 1}{x-1} ; \; a = 1 \)

 

5. Find the numbers at which \( f \) is discontinuous. Show work.   solution

 \( f(x)=\begin{cases}  1+x^2, \quad  x \le 0 \\ 2-x, \quad  0 < x  \le 2 \\ (x-2)^2, \quad x >  2 \end{cases}\)

6. The value of \( k \) so that the function \( f(x) \) is continuous in \((-\infty, \infty)\), where \(f(x) =\begin{cases} 5, \quad x < 1 \\ 2k+x, \quad x\geq 1\end{cases}\) .

solution

 

7. Find the values of \(a \) and \(b\) that make \(f\)continuous everywhere.    \( \displaystyle{ f(x)=\begin{cases} \dfrac{x^{2}-4}{x-2} \quad \text { if } \quad x < 2 \\ ax^2-bx+3 \quad\text { if } \quad 2 \leqslant x < 3 \\ 2x-a+b \quad \text { if }\quad x \geqslant 3 \end{cases}}\)    solution

8. Evaluate the limits:    solution

(a) \(\displaystyle{\lim_{x \to 1/2} \tan (\pi x /2)} \)      (b) \(\displaystyle{\lim_{x \to 0} e^{-\cos x}} \)     (c) \(\displaystyle{\lim_{x \to   1/2}\sin\left(\frac{\pi}{2}\,{x}\right)}\)

 

9. Given that \( f \) is continuous from left at  \(  x = 0\) but discontinuous at \(x = 0 \). Which of the following must be true?

I. \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) \)

 II. \( \lim_{x \to 0^-} f(x)=f(0) \)

III.\( \lim_{x \to 0^+} f(x)=f(0) \)

solution