Limits & Continuity - IB Mathematics Analysis and Approaches

1. Let $f(x) = \frac{1}{x-1}$ and $g(x) = x^2$. The composite function $h(x) = f(g(x))$ is discontinuous at which value(s) of $x$?
   Answer: At $x=-1$ and $x=1$. When dealing with composite functions and discontinuities, you need to find where the outer function becomes undefined after substitution. A function is discontinuous where it's not defined or where there's a break in the graph. First, let's find the composite function $h(x) = f(g(x))$. Since $f(x) = \frac{1}{x-1}$ and $g(x) = x^2$, we substitute $g(x)$ into $f(x)$: $h(x) = f(g(x)) = f(x^2) = \frac{1}{x^2-1}$ Now, $h(x)$ is discontinuous wherever the denominator equals zero. Setting $x^2 - 1 = 0$: $x^2 = 1$ $x = \pm 1$ So $h(x)$ is discontinuous at $x = -1$ and $x = 1$. Looking at the answer choices: A correctly identifies both discontinuities at $x = -1$ and $x = 1$. B incorrectly includes $x = 0$, which would make the denominator $0^2 - 1 = -1 \neq 0$, so the function is actually defined there. C only identifies $x = 1$, missing the discontinuity at $x = -1$. D claims there are no discontinuities, which ignores the fact that rational functions are undefined when their denominators equal zero. The correct answer is A. Study tip: For composite function discontinuities, always substitute completely first, then find where the result is undefined. Don't just look at where the individual functions are discontinuous—the composition can create new discontinuities or eliminate existing ones.
2. A function $f(x)$ is continuous at a point $x=c$. Which of the following conditions is NOT necessarily required by the definition of continuity at $x=c$?
   Answer: $f(x)$ is differentiable at $x=c$. The definition of continuity for a function $f(x)$ at a point $x=c$ requires three conditions: 1) $f(c)$ must be defined, 2) $\lim_{x \to c} f(x)$ must exist, and 3) $\lim_{x \to c} f(x) = f(c)$. Differentiability at a point implies continuity, but continuity does not imply differentiability. For example, the function $f(x) = |x|$ is continuous at $x=0$ but it is not differentiable at $x=0$. Therefore, differentiability is not a necessary condition for continuity.
3. What is the value of $\lim_{x \to 2^+} \frac{5}{x-2}$?
   Answer: $\infty$ The notation $x \to 2^+$ means that $x$ approaches 2 from the right side, so $x$ is a number slightly greater than 2. This means the denominator, $x-2$, is a small positive number. The numerator is a positive constant, 5. Dividing a positive constant by a small positive number results in a very large positive number. Therefore, the limit is $\infty$.
4. Find the value of the limit $\lim_{x \to 3} \frac{x^2 - x - 6}{x - 3}$.
   Answer: 5 To evaluate this limit, we first try to substitute $x=3$, which results in the indeterminate form $\frac{0}{0}$. This suggests that $(x-3)$ is a factor of the numerator. Factoring the numerator gives $x^2 - x - 6 = (x-3)(x+2)$. The expression becomes $\lim_{x \to 3} \frac{(x-3)(x+2)}{x - 3}$. For $x \neq 3$, we can cancel the $(x-3)$ terms, leaving $\lim_{x \to 3} (x+2)$. Now we can substitute $x=3$ to get $3+2 = 5$.
5. Find the value of the limit $\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$.
   Answer: $\frac{1}{6}$ Substituting $x=9$ gives the indeterminate form $\frac{0}{0}$. We can resolve this by multiplying the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x} + 3$. $\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \lim_{x \to 9} \frac{(\sqrt{x})^2 - 3^2}{(x - 9)(\sqrt{x} + 3)} = \lim_{x \to 9} \frac{x - 9}{(x - 9)(\sqrt{x} + 3)}$. We can cancel the $(x-9)$ terms, leaving $\lim_{x \to 9} \frac{1}{\sqrt{x} + 3}$. Substituting $x=9$ now gives $\frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$.
6. Let $f(x) = \lfloor x \rfloor$ be the greatest integer function (floor function). Which statement correctly describes the function at $x=3$?
   Answer: $\lim_{x \to 3^-} f(x) = 2$ and $\lim_{x \to 3^+} f(x) = 3$. When analyzing the floor function $f(x) = \lfloor x \rfloor$ at integer values, you need to examine one-sided limits since the function has jump discontinuities at these points. The floor function outputs the greatest integer less than or equal to x. At $x = 3$, we have $f(3) = 3$. However, the behavior as we approach 3 from different sides tells the full story. For the left-hand limit $\lim_{x \to 3^-} f(x)$: As x approaches 3 from the left (values like 2.9, 2.99, 2.999), we're looking at numbers in the interval $[2,3)$. For all these values, $\lfloor x \rfloor = 2$. Therefore, $\lim_{x \to 3^-} f(x) = 2$. For the right-hand limit $\lim_{x \to 3^+} f(x)$: As x approaches 3 from the right (values like 3.1, 3.01, 3.001), we're in the interval $(3,4)$. For all these values, $\lfloor x \rfloor = 3$. Therefore, $\lim_{x \to 3^+} f(x) = 3$. This confirms answer A is correct. Answer B is wrong because $\lim_{x \to 3} f(x)$ doesn't exist - the one-sided limits aren't equal (2 ≠ 3). Answer C is wrong because continuity requires the limit to exist and equal the function value. Since the limit doesn't exist, the function is discontinuous at $x = 3$. Answer D incorrectly states both one-sided limits, confusing the intervals where the floor function takes different values. Study tip: For floor function problems, always sketch the step graph and identify which integer interval contains your approaching values.
7. What is the value of $\lim_{x \to 4} (h(x) + x^2)$?
   Answer: 15 By the definition of continuity, if $h(x)$ is continuous at $x=4$, then $\lim_{x \to 4} h(x) = h(4)$. We are given $h(4) = -1$, so $\lim_{x \to 4} h(x) = -1$. Using the properties of limits, the limit of a sum is the sum of the limits: $\lim_{x \to 4} (h(x) + x^2) = \lim_{x \to 4} h(x) + \lim_{x \to 4} x^2$. We can evaluate these limits: $-1 + 4^2 = -1 + 16 = 15$.
8. Which of the following statements about the function $f$ is true?
   Answer: $\lim_{x \to 1^-} f(x) = -2$. We must analyze the function by considering the definition of $|x-1|$. \\ Case 1: $x > 1$. Then $|x-1| = x-1$. So, $f(x) = \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1$. The right-hand limit is $\lim_{x \to 1^+} (x+1) = 1+1 = 2$. \\ Case 2: $x < 1$. Then $|x-1| = -(x-1)$. So, $f(x) = \frac{x^2-1}{-(x-1)} = \frac{(x-1)(x+1)}{-(x-1)} = -(x+1)$. The left-hand limit is $\lim_{x \to 1^-} -(x+1) = -(1+1) = -2$. \\ Since the left-hand limit (-2) is not equal to the right-hand limit (2), the overall limit $\lim_{x \to 1} f(x)$ does not exist. This means the function has a jump discontinuity at $x=1$ and is not continuous there. Option D correctly states the value of the left-hand limit.
9. Find the value of $\lim_{x \to \infty} \frac{6x^3 - 4x + 1}{8x + 2x^3}$.
   Answer: 3 To find the limit of a rational function as $x \to \infty$, we compare the degrees of the numerator and the denominator. The degree of the numerator is 3, and the degree of the denominator is also 3. Since the degrees are equal, the limit is the ratio of the leading coefficients. The leading term in the numerator is $6x^3$ and the leading term in the denominator is $2x^3$. The limit is $\frac{6}{2} = 3$. Alternatively, divide every term by the highest power of $x$, which is $x^3$: $\lim_{x \to \infty} \frac{6 - \frac{4}{x^2} + \frac{1}{x^3}}{\frac{8}{x^2} + 2}$. As $x \to \infty$, the terms with $x$ in the denominator approach 0, leaving $\frac{6-0+0}{0+2} = \frac{6}{2} = 3$.
10. Find $\lim_{x \to -\infty} \frac{3x^2 + 5x}{1 - 2x^2}$.
    Answer: -$\frac{3}{2}$ To find the limit of a rational function as $x \to -\infty$, we can compare the degrees of the numerator and denominator. Both have a degree of 2. Therefore, the limit is the ratio of the leading coefficients. The leading term in the numerator is $3x^2$, and the leading term in the denominator is $-2x^2$. The limit is $\frac{3}{-2} = -\frac{3}{2}$. The direction $x \to -\infty$ does not change this result compared to $x \to \infty$ when the degrees are equal.
11. Given that $\lim_{x \to a^-} f(x) = L$ and $\lim_{x \to a^+} f(x) = M$, where $L \neq M$. Which of the following must be true?
    Answer: $\lim_{x \to a} f(x)$ does not exist. The condition for the existence of a limit at a point $x=a$ is that the left-hand limit must equal the right-hand limit. The problem states that $L \neq M$, meaning $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$. Therefore, the overall limit $\lim_{x \to a} f(x)$ does not exist. This describes a jump discontinuity. A removable discontinuity requires the limit to exist. It is possible for $f(a)$ to be defined (e.g., $f(a)=L$ or $f(a)=M$), so B is not necessarily true. This situation describes a jump, not a vertical asymptote, so D is incorrect.
12. Which of the following statements is true about the function $f$ at $x=2$?
    Answer: $f(x)$ is continuous at $x=2$. To check for continuity at $x=2$, we must check three conditions: 1. $f(2)$ is defined. 2. $\lim_{x \to 2} f(x)$ exists. 3. $\lim_{x \to 2} f(x) = f(2)$. \\ First, $f(2) = 2(2) + 1 = 5$. \\ Second, we check the one-sided limits. The left-hand limit is $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2+1) = 2^2+1 = 5$. The right-hand limit is $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x+1) = 2(2)+1 = 5$. Since the left-hand and right-hand limits are equal, $\lim_{x \to 2} f(x) = 5$. \\ Finally, since $\lim_{x \to 2} f(x) = 5 = f(2)$, the function is continuous at $x=2$.
13. The function $f(x) = \frac{x^2 - 4}{x+2}$ has a discontinuity at $x=-2$. What type of discontinuity is it?
    Answer: A removable discontinuity. The function is undefined at $x=-2$. To determine the type of discontinuity, we evaluate the limit as $x$ approaches -2. We can factor the numerator: $f(x) = \frac{(x-2)(x+2)}{x+2}$. For $x \neq -2$, we can cancel the $(x+2)$ term, so $f(x) = x-2$. The limit is $\lim_{x \to -2} (x-2) = -2 - 2 = -4$. Since the limit exists but the function is not defined at $x=-2$, the discontinuity is removable. The graph is a line with a hole at the point $(-2, -4)$.
14. Which theorem guarantees that there is a value $c$ in the interval $(-2, 5)$ such that $f(c)=0$?
    Answer: The Intermediate Value Theorem The Intermediate Value Theorem (IVT) states that if a function $f$ is continuous on a closed interval $[a, b]$, then for any value $N$ between $f(a)$ and $f(b)$, there exists at least one number $c$ in $(a, b)$ such that $f(c) = N$. In this case, the function is continuous on $[-2, 5]$, $f(-2)=3$, and $f(5)=-1$. Since 0 is a value between -1 and 3, the IVT guarantees that there is a value $c$ in $(-2, 5)$ such that $f(c)=0$.
15. A function $g(x)$ has a vertical asymptote at $x=1$ and a horizontal asymptote at $y=-3$. Which of the following statements must be true based on this information?
    Answer: $\lim_{x \to \infty} g(x) = -3$ The definition of a horizontal asymptote at $y=L$ is that $\lim_{x \to \infty} g(x) = L$ or $\lim_{x \to -\infty} g(x) = L$. Given the horizontal asymptote is at $y=-3$, it must be that at least one of these limits is -3. Option B matches this definition. A vertical asymptote at $x=1$ means $\lim_{x \to 1} g(x)$ approaches $\pm\infty$, so A is incorrect. A vertical asymptote at $x=1$ implies $g(1)$ is undefined, so D is incorrect. C incorrectly swaps the roles of $x$ and $y$ in the limit definition.
16. For which value of $a$ is the function $f(x) = \begin{cases} ae^x + 1 & \text{if } x \le 0 \\ x^2 + 3 & \text{if } x > 0 \end{cases}$ continuous at $x=0$?
    Answer: 2 For continuity at $x=0$, the left-hand limit must equal the right-hand limit. We calculate both limits. \\ Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (ae^x + 1) = a e^0 + 1 = a(1) + 1 = a+1$. \\ Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 + 3) = 0^2 + 3 = 3$. \\ Set the limits equal to each other: $a+1 = 3$. Solving for $a$ gives $a=2$.
17. Find the value of $\lim_{x \to 0} \frac{\sin(3x)}{\sin(x)}$.
    Answer: 3 This limit is of the indeterminate form $\frac{0}{0}$. We can use the known limit $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$. We can rewrite the expression by multiplying the numerator and denominator by $3x$ and $x$ respectively: $\lim_{x \to 0} \frac{\sin(3x)}{\sin(x)} = \lim_{x \to 0} \left( \frac{\sin(3x)}{3x} \cdot 3x \cdot \frac{x}{\sin(x)} \cdot \frac{1}{x} \right) = \lim_{x \to 0} \left( \frac{\sin(3x)}{3x} \cdot \frac{x}{\sin(x)} \cdot \frac{3x}{x} \right)$. As $x \to 0$, $3x \to 0$, so $\frac{\sin(3x)}{3x} \to 1$. Also, $\frac{x}{\sin(x)} \to 1$. The expression simplifies to $1 \cdot 1 \cdot 3 = 3$.

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