Section 2.5: Continuity

(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 \(x\).

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

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

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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 \(f(x) =\begin{cases} 5, \quad x < 1 \\ 2k+x, \quad x\geq 1\end{cases}\) .

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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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