Continuity at a Point - Calculus 1

1. Let $f(x) = \frac{1}{x-2}$. The Intermediate Value Theorem states that for a continuous function on $[a, b]$, the function takes on every value between $f(a)$ and $f(b)$. Since $f(1)=-1$ and $f(3)=1$, why does the theorem not guarantee a root $c$ in $(1,3)$ where $f(c)=0$?
   Answer: The function is not continuous on the interval $[1,3]$. The Intermediate Value Theorem has a crucial precondition: the function must be continuous on the closed interval $[a, b]$. The function $f(x) = \frac{1}{x-2}$ has an infinite discontinuity at $x=2$, which is within the interval $[1,3]$. Because the function is not continuous on this interval, the theorem's conclusion is not guaranteed.
2. Let $f(x) = \begin{cases} a\cos(x) & x \le 0 \\ x^2+2a-3 & x > 0 \end{cases}$. For what value of $a$ is $f$ continuous at $x=0$?
   Answer: $a = 3$ For continuity at $x=0$, we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$. Let's calculate the limits and the function value. $f(0) = a\cos(0) = a \cdot 1 = a$. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} a\cos(x) = a\cos(0) = a$. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2+2a-3) = 0^2+2a-3 = 2a-3$. Equating the left and right limits: $a = 2a-3$. Solving for $a$ gives $-a = -3$, so $a=3$.
3. If a function $f$ is differentiable at a point $x=c$, which of the following statements about $f$ at $x=c$ must be true?
   Answer: $f$ is continuous. A fundamental theorem in calculus states that if a function is differentiable at a point, it must be continuous at that point. Differentiability is a stronger condition than continuity. If the derivative exists, the function cannot have any type of discontinuity (removable, jump, or infinite) at that point.
4. The function $f(x) = \frac{x^2+ax-10}{x-2}$ has a removable discontinuity at $x=2$. What is the value of the constant $a$?
   Answer: $a = 3$ For $f(x)$ to have a removable discontinuity at $x=2$, the limit $\lim_{x \to 2} f(x)$ must exist. This requires the numerator to be zero at $x=2$, so that the factor $(x-2)$ can be cancelled. $(2)^2 + a(2) - 10 = 0$ $4 + 2a - 10 = 0$ $2a - 6 = 0$ $a = 3$ If $a=3$, then $f(x) = \frac{x^2+3x-10}{x-2} = \frac{(x-2)(x+5)}{x-2} = x+5$ for $x \neq 2$. The limit as $x \to 2$ is 7, so it is a removable discontinuity.
5. Consider the function $f(x) = \begin{cases} \frac{x-1}{|x-1|} & x \neq 1 \\ 1 & x=1 \end{cases}$. Which statement accurately describes the continuity of $f(x)$ at $x=1$?
   Answer: The function is continuous from the right at $x=1$. When analyzing continuity at a point, you need to check three conditions: the function must be defined at the point, the limit must exist, and the limit must equal the function value. For piecewise functions involving absolute values, examining one-sided limits is crucial. Let's analyze $f(x) = \begin{cases} \frac{x-1}{|x-1|} & x \neq 1 \\ 1 & x=1 \end{cases}$ at $x=1$. First, simplify $\frac{x-1}{|x-1|}$ by considering the sign of $(x-1)$. When $x > 1$, we have $|x-1| = x-1$, so $\frac{x-1}{|x-1|} = \frac{x-1}{x-1} = 1$. When $x < 1$, we have $|x-1| = -(x-1)$, so $\frac{x-1}{|x-1|} = \frac{x-1}{-(x-1)} = -1$. Now check the one-sided limits: - Right-hand limit: $\lim_{x \to 1^+} f(x) = 1$ - Left-hand limit: $\lim_{x \to 1^-} f(x) = -1$ - Function value: $f(1) = 1$ Since the right-hand limit equals the function value at $x=1$, option A is correct - the function is continuous from the right. Option B is wrong because removable discontinuities occur when limits exist but don't equal function values. Option C is wrong because full continuity requires both one-sided limits to exist and be equal. Option D is wrong because the left-hand limit (-1) doesn't equal $f(1) = 1$. Study tip: For absolute value functions, always split into cases based on where the expression inside changes sign, then check one-sided limits separately.
6. How should $f(0)$ be defined to make the function $f(x) = \frac{(1+x)^3 - 1}{x}$ continuous at $x=0$?
   Answer: $f(0) = 3$ For $f$ to be continuous at $x=0$, we must define $f(0)$ to be equal to the limit of $f(x)$ as $x$ approaches 0. We can find this limit by expanding the numerator: $\lim_{x \to 0} \frac{(1+x)^3 - 1}{x} = \lim_{x \to 0} \frac{(1 + 3x + 3x^2 + x^3) - 1}{x}$ $= \lim_{x \to 0} \frac{3x + 3x^2 + x^3}{x}$ $= \lim_{x \to 0} (3 + 3x + x^2)$ $= 3 + 3(0) + (0)^2 = 3$ Therefore, we should define $f(0) = 3$.
7. The function $f(x) = \begin{cases} x & \text{if } x \text{ is rational} \\ 2-x & \text{if } x \text{ is irrational} \end{cases}$ is continuous at which of the following points?
   Answer: $x=1$ For the function to be continuous at a point $c$, the limit must exist. This can only happen if the values of the two pieces of the function approach each other as $x$ approaches $c$. Therefore, we must have $c = 2-c$. Solving this equation gives $2c = 2$, so $c = 1$. At $x=1$, both pieces of the function evaluate to 1. For any sequence of points approaching 1, rational or irrational, the function values will approach 1. So, $\lim_{x \to 1} f(x) = 1$. Since $1$ is rational, $f(1)=1$. The limit equals the function value, so $f$ is continuous at $x=1$ only.
8. The function $f(x) = \begin{cases} \frac{e^{2x}-1}{x}, & x \neq 0 \\ k, & x=0 \end{cases}$ is continuous at $x=0$. What is the value of $k$?
   Answer: $k=2$ For $f$ to be continuous at $x=0$, $k$ must be equal to $\lim_{x \to 0} f(x)$. We need to evaluate $\lim_{x \to 0} \frac{e^{2x}-1}{x}$. This limit is of the indeterminate form $0/0$. We can use L'Hôpital's Rule. $\lim_{x \to 0} \frac{\frac{d}{dx}(e^{2x}-1)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{2e^{2x}}{1} = 2e^{2(0)} = 2e^0 = 2(1) = 2$ Alternatively, this is the definition of the derivative of $g(x)=e^{2x}$ at $x=0$. $g'(x)=2e^{2x}$, so $g'(0)=2$. Thus, $k=2$.
9. Let $f(x) = x-2$ and let $g(x)$ be a piecewise function defined as $g(x) = \begin{cases} -1 & \text{if } x < 2 \\ 1 & \text{if } x \ge 2 \end{cases}$. If $h(x) = f(x)g(x)$, which of the following statements is true about the continuity of $h(x)$ at $x=2$?
   Answer: $h(x)$ is continuous at $x=2$. To determine if $h(x) = f(x)g(x)$ is continuous at $x=2$, we check the three conditions for continuity. 1. Check if $h(2)$ is defined: $h(2) = f(2)g(2) = (2-2)(1) = 0$. It is defined. 2. Check if $\lim_{x \to 2} h(x)$ exists. We evaluate the one-sided limits: Left-hand limit: $\lim_{x \to 2^-} h(x) = \lim_{x \to 2^-} (x-2)g(x) = \lim_{x \to 2^-} (x-2)(-1) = (2-2)(-1) = 0$. Right-hand limit: $\lim_{x \to 2^+} h(x) = \lim_{x \to 2^+} (x-2)g(x) = \lim_{x \to 2^+} (x-2)(1) = (2-2)(1) = 0$. Since the left-hand and right-hand limits are equal, the limit exists and $\lim_{x \to 2} h(x) = 0$. 3. Check if $\lim_{x \to 2} h(x) = h(2)$. We found that both are equal to 0. Since all three conditions are satisfied, $h(x)$ is continuous at $x=2$. The discontinuity of $g(x)$ is 'cancelled out' because $f(x)$ approaches 0 at $x=2$.
10. The function $f(x) = \begin{cases} \frac{\tan(kx)}{x} & x \neq 0 \\ 4 & x=0 \end{cases}$ is continuous at $x=0$. What is the value of $k$?
    Answer: $k=4$ For continuity at $x=0$, we need $\lim_{x \to 0} f(x) = f(0)$. We are given $f(0)=4$. We evaluate the limit: $$\lim_{x \to 0} \frac{\tan(kx)}{x} = \lim_{x \to 0} \frac{\sin(kx)}{x \cos(kx)}$$ $$= \lim_{x \to 0} \left( \frac{\sin(kx)}{kx} \cdot k \cdot \frac{1}{\cos(kx)} \right)$$ As $x \to 0$, $\frac{\sin(kx)}{kx} \to 1$ and $\cos(kx) \to \cos(0) = 1$. So the limit is: $$(1) \cdot k \cdot \frac{1}{1} = k$$ Setting the limit equal to the function value: $k=4$.
11. Let $f(x) = \lfloor x^2 \rfloor$. At which of the following points is $f(x)$ continuous?
    Answer: $x = 1.5$ The greatest integer function $\lfloor u \rfloor$ is discontinuous whenever its argument $u$ is an integer. Therefore, $f(x) = \lfloor x^2 \rfloor$ is discontinuous whenever $x^2$ is an integer. Let's check the options: A) If $x=\sqrt{2}$, $x^2=2$, which is an integer. Discontinuous. B) If $x=\sqrt{3}$, $x^2=3$, which is an integer. Discontinuous. C) If $x=2$, $x^2=4$, which is an integer. Discontinuous. D) If $x=1.5$, $x^2=2.25$, which is not an integer. The function is continuous at this point because for values of $x$ close to 1.5 (e.g., in the interval (1.4, 1.6)), $x^2$ will be close to 2.25, and $\lfloor x^2 \rfloor$ will remain constant at 2.
12. Let the function $f$ be defined as $f(x) = \begin{cases} x^k \cos(\frac{\pi}{x}) & \text{if } x \neq 0 \\ L & \text{if } x = 0 \end{cases}$. If $f$ is continuous at $x=0$, what must be true about the constants $k$ and $L$?
    Answer: $k > 0$ and $L = 0$ For $f$ to be continuous at $x=0$, we must have $\lim_{x \to 0} f(x) = f(0)$. Here, $f(0) = L$. So we need to evaluate $\lim_{x \to 0} x^k \cos(\frac{\pi}{x})$. We can use the Squeeze Theorem. The cosine function is bounded: $-1 \le \cos(\frac{\pi}{x}) \le 1$ for all $x \neq 0$. Multiplying by $x^k$ (assuming $x^k > 0$, we can use $|x^k|$ for the general case): $-|x^k| \le x^k \cos(\frac{\pi}{x}) \le |x^k|$. For the Squeeze Theorem to imply that the limit is 0, we need the limits of the outer functions to be 0. That is, we need $\lim_{x \to 0} |x^k| = 0$. This condition holds if and only if $k > 0$. If $k>0$, then $\lim_{x \to 0} x^k \cos(\frac{\pi}{x}) = 0$. For continuity, we must have $L = \lim_{x \to 0} f(x)$, so $L=0$. Therefore, we need $k>0$ and $L=0$.
13. If $\lim_{x \to c^-} f(x) = L$, $\lim_{x \to c^+} f(x) = M$, and $f(c) = L$, under which condition is $f$ continuous at $x=c$?
    Answer: Only if $L = M$ For a function to be continuous at $x=c$, three conditions must be met: 1) $f(c)$ is defined, 2) $\lim_{x \to c} f(x)$ exists, and 3) $\lim_{x \to c} f(x) = f(c)$. The two-sided limit exists if and only if the left-hand limit equals the right-hand limit, i.e., $L=M$. If $L=M$, then the limit is $L$. Since we are given that $f(c)=L$, all three conditions for continuity would be satisfied if $L=M$.
14. Suppose for a function $f$ and a real number $c$, it is known that $\lim_{x \to c} f(x) = L$, where $L$ is a finite real number. Which of the following additional conditions is both necessary and sufficient to guarantee that $f$ is continuous at $x=c$?
    Answer: The value of $f(c)$ must be equal to $L$. The definition of continuity at a point $c$ requires three conditions: (1) $f(c)$ is defined, (2) $\lim_{x \to c} f(x)$ exists, and (3) $\lim_{x \to c} f(x) = f(c)$. The stem states that condition (2) is met, with the limit being $L$. Choice A, $f(c)$ is defined, is necessary for continuity, but not sufficient. For example, $f(c)$ could be defined as $L+1$, in which case condition (3) would fail. Choice B, $f(c)=L$, combines conditions (1) and (3). If $f(c)=L$, then $f(c)$ is defined and it equals the limit. This is precisely the third condition required for continuity and is therefore both necessary and sufficient, given that the limit exists. Choice C, differentiability at $c$, is a sufficient condition for continuity, but it is not necessary. A function can be continuous without being differentiable (e.g., $f(x)=|x|$ at $x=0$). Choice D is incorrect; there is no general requirement that $f(c)$ must be 0 for continuity.
15. Let $f(x) = \begin{cases} x^2+1 & x \le c \\ 3x-1 & x > c \end{cases}$. For what value of $c$ is the function $f$ continuous everywhere?
    Answer: $c = 1$ or $c = 2$ The function pieces are polynomials, so they are continuous everywhere except possibly at the point $x=c$ where the definition changes. For continuity at $x=c$, the left-hand limit must equal the right-hand limit. $$\lim_{x \to c^-} (x^2+1) = \lim_{x \to c^+} (3x-1)$$ $$c^2+1 = 3c-1$$ Rearranging gives a quadratic equation: $$c^2 - 3c + 2 = 0$$ Factoring the quadratic: $$(c-1)(c-2) = 0$$ This yields two possible values for $c$: $c=1$ and $c=2$. For either of these values, the function $f$ will be continuous everywhere.
16. Let $f(x) = \frac{x^2 + 2x - 8}{x^3 - 8}$. The function $f$ is discontinuous at $x=2$. Which of the following actions would make $f$ continuous at $x=2$?
    Answer: Define $f(2) = \frac{1}{2}$. A function has a removable discontinuity at a point $c$ if $\lim_{x \to c} f(x)$ exists but is not equal to $f(c)$ (or $f(c)$ is undefined). To make the function continuous, we must define $f(c)$ to be equal to the limit. First, find the limit of $f(x)$ as $x \to 2$. We factor the numerator and denominator: Numerator: $x^2 + 2x - 8 = (x+4)(x-2)$. Denominator (difference of cubes): $x^3 - 8 = (x-2)(x^2 + 2x + 4)$. So, for $x \neq 2$, $f(x) = \frac{(x+4)(x-2)}{(x-2)(x^2 + 2x + 4)} = \frac{x+4}{x^2 + 2x + 4}$. Now we can evaluate the limit: $\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x+4}{x^2 + 2x + 4} = \frac{2+4}{2^2 + 2(2) + 4} = \frac{6}{4+4+4} = \frac{6}{12} = \frac{1}{2}$. Since the limit exists, the discontinuity is removable. To make $f$ continuous at $x=2$, we must define $f(2)$ to be the value of the limit. Thus, we should define $f(2) = \frac{1}{2}$.
17. Let $f(x) = \begin{cases} \frac{\sqrt{x+1}-2}{x-3} & x \neq 3 \\ c & x=3 \end{cases}$. For what value of $c$ is $f$ continuous at $x=3$?
    Answer: $c = \frac{1}{4}$ For continuity at $x=3$, $c$ must equal the limit of the function as $x \to 3$. To find the limit, we multiply by the conjugate of the numerator: $$\lim_{x \to 3} \frac{\sqrt{x+1}-2}{x-3} \cdot \frac{\sqrt{x+1}+2}{\sqrt{x+1}+2}$$ $$= \lim_{x \to 3} \frac{(x+1)-4}{(x-3)(\sqrt{x+1}+2)}$$ $$= \lim_{x \to 3} \frac{x-3}{(x-3)(\sqrt{x+1}+2)}$$ $$= \lim_{x \to 3} \frac{1}{\sqrt{x+1}+2} = \frac{1}{\sqrt{3+1}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4}$$ Therefore, $c = \frac{1}{4}$.
18. Let $f(x) = \begin{cases} \frac{\sin(3x)}{x} & x < 0 \\ a & x = 0 \\ \frac{x^2-bx}{x} & x > 0 \end{cases}$. For which values of $a$ and $b$ is $f(x)$ continuous at $x=0$?
    Answer: $a = 3, b = -3$ For continuity at $x=0$, the left-hand limit, right-hand limit, and function value must all be equal. 1. Left-hand limit: $\lim_{x \to 0^-} \frac{\sin(3x)}{x} = \lim_{x \to 0^-} 3 \cdot \frac{\sin(3x)}{3x} = 3 \cdot 1 = 3$. 2. Function value: $f(0) = a$. 3. Right-hand limit: $\lim_{x \to 0^+} \frac{x^2-bx}{x} = \lim_{x \to 0^+} (x-b) = 0-b = -b$. Setting them all equal: $3 = a = -b$. This gives $a = 3$ and $b = -3$.
19. Let $f$ be a function such that $\lim_{x \to c} f(x) = L$, where $L$ is a finite number. Which of the following statements must be true to ensure that $f$ is continuous at $x=c$?
    Answer: $f(c)$ must be defined and $f(c)=L$. The definition of continuity at a point $c$ has three conditions: 1) $f(c)$ is defined, 2) $\lim_{x \to c} f(x)$ exists, and 3) $\lim_{x \to c} f(x) = f(c)$. The stem already states that condition 2 is met (the limit exists and equals $L$). Therefore, to ensure continuity, conditions 1 and 3 must also be met, which are combined in the statement $f(c)=L$.
20. Let $f(x) = x - \lfloor x \rfloor$, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. Which statement best describes the continuity of $f$ at $x=3$?
    Answer: $f$ has a jump discontinuity at $x=3$ because the one-sided limits are not equal. We test the three conditions for continuity at $x=3$. 1. Function value: $f(3) = 3 - \lfloor 3 \rfloor = 3 - 3 = 0$. The function is defined at $x=3$. 2. Left-hand limit: As $x$ approaches 3 from the left, $x$ is slightly less than 3 (e.g., 2.999). For such values, $\lfloor x \rfloor = 2$. So, $\lim_{x \to 3^-} (x - \lfloor x \rfloor) = \lim_{x \to 3^-} (x - 2) = 3 - 2 = 1$. 3. Right-hand limit: As $x$ approaches 3 from the right, $x$ is slightly greater than 3 (e.g., 3.001). For such values, $\lfloor x \rfloor = 3$. So, $\lim_{x \to 3^+} (x - \lfloor x \rfloor) = \lim_{x \to 3^+} (x - 3) = 3 - 3 = 0$. Since the left-hand limit (1) is not equal to the right-hand limit (0), the overall limit $\lim_{x \to 3} f(x)$ does not exist. This is the definition of a jump discontinuity.

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