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 \( a=5\) View Solution (b) \(\displaystyle{ f(x)=\begin{cases} x+1, & x \le 2 \\ 2x-1, & x > 2 \end{cases}}\)   at \( a=2\) View Solution
2. Consider the following graph of \(f\). Is \( f \) continuous at \( x = -2, 0, 2, 3\)? Justify your answer using the definition of continuity or state the type of discontinuity. View Solution
3. Explain why the function \(f\) is discontinuous at \( 1 \). View Solution \( f(x)=\begin{cases} \dfrac{2x-2}{x^2-1}, & x \ne 1 \\ 2, & x = 1 \end{cases}\)
4. Determine whether the function is continuous at \( x = a \). If discontinuous, state the type of discontinuity. View 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. View Solution \( f(x)=\begin{cases} 1+x^2, & x \le 0 \\ 2-x, & 0 < x \le 2 \\ (x-2)^2, & x > 2 \end{cases}\)
6. Find the value of \( k \) so that the function \( f(x) \) is continuous in \((-\infty, \infty)\). View Solution \(f(x) =\begin{cases} 5, & x < 1 \\ 2k+x, & x \geq 1\end{cases}\)
7. Find the values of \(a \) and \(b\) that make \(f\) continuous everywhere. View Solution \( \displaystyle{ f(x)=\begin{cases} \dfrac{x^{2}-4}{x-2} & \text{if } x < 2 \\ ax^2-bx+3 & \text{if } 2 \leqslant x < 3 \\ 2x-a+b & \text{if } x \geqslant 3 \end{cases}}\)
8. Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. View Solution \( 4x^3-6x^2+3x-2=0 \) on the interval \( (1, 2) \)
9. Use the Intermediate Value Theorem to show that there is a zero of the function in the given interval. View Solution \( f(x)=x^2-2x-2 \) on the interval \( (-1, 0) \)
10. Use the IVT to show that the equation \(e^{-x} = x \) has a solution in the interval \( (0, 1)\). View Solution
11. Evaluate the following limits: View Solution (a) \(\displaystyle{\lim_{x \to 1/2} \tan (\pi x /2)} \)   (b) \(\displaystyle{\lim_{x \to 0} e^{-\cos x}} \)   (c) \(\displaystyle{\lim_{x \to 1/2}\sin\left(\frac{\pi}{2}\,{x}\right)}\)
12. Given that \( f \) is continuous from the left at \( x = 0 \) but discontinuous at \( x = 0 \), which must be true? View Solution I. \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) \)
    II. \( \lim_{x \to 0^-} f(x)=f(0) \)
    III. \( \lim_{x \to 0^+} f(x)=f(0) \)