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](ex11.png)
3. Explain why the function \(f\) is discontinuous at \( 1 \).
\( f(x)=\begin{cases} \dfrac{2x-2}{x^2-1}, \:\: x \ne 1 \\ 2, \:\: x = 1. \end{cases}\)
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. Find thhe value of \( k \) so that the function \( f(x) \) is continuous
in \((-\infty, \infty)\), where \(f(x) =\begin{cases} 5, \quad
x < 1 \\ 2k+x, \quad x\geq 1\end{cases}\) .
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
7. Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. \( 4x^3-6x^2+3x-2=0,
\ (1, 2) \).
solution
8. Use the Intermediate Value Theorem to show that there is a zero of the function \(f(x)=x^2-2x-2 \) in the interval
\( (-1, 0) \). solution
9. Use the IVT to show that the equation \(e^{-x} = x \) has a
solution in the interval \( (0, 1)\).
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.\( \lim_{x \to 0^+} f(x)=f(0) \)
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