4.1 MCQs-Limits and Continuity

Limits and Continuity MCQs

Concept of Limits

1. The statement $\lim_{x \to a} f(x) = L$ means that:

1. $f(a) = L$

2. $f(x)$ gets arbitrarily close to L as x gets sufficiently close to a (from either side)

3. $f(x)$ equals L for all x near a

4. The value of $f(a)$ exists

Show me the answer

Answer: 2. $f(x)$ gets arbitrarily close to L as x gets sufficiently close to a (from either side)

Explanation:

• The limit describes the behavior of the function $f(x)$ as x approaches a, not necessarily the value at $x = a$.

• The function may or may not be defined at $x = a$, and even if defined, $f(a)$ may not equal L.

• The formal ($\epsilon-\delta$) definition captures this idea of making $f(x)$ arbitrarily close to L by making x sufficiently close to a.

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2. $\lim_{x \to 0} \frac{\sin x}{x}$ is equal to:

1. 0

2. 1

3. Does not exist

4. $\infty$

Show me the answer

Answer: 2. 1

Explanation:

• This is a standard trigonometric limit.

• $\lim_{x \to 0} \frac{\sin x}{x} = 1$

• This result is fundamental and is often proved using the Squeeze Theorem or geometric arguments.

• Note that $\lim_{x \to 0} \frac{\tan x}{x} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ are also important.

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3. If $\lim_{x \to a^-} f(x) = 5$ and $\lim_{x \to a^+} f(x) = 5$, then:

1. $\lim_{x \to a} f(x) = 5$

2. $f(a) = 5$

3. $f(x)$ is continuous at $x = a$

4. The limit does not exist

Show me the answer

Answer: 1. $\lim_{x \to a} f(x) = 5$

Explanation:

• For the (two-sided) limit $\lim_{x \to a} f(x)$ to exist and equal L, both the left-hand limit ($x \to a^-$) and the right-hand limit ($x \to a^+$) must exist and be equal to L.

• Here, both one-sided limits equal 5, so the two-sided limit exists and equals 5.

• This does not guarantee that $f(a) = 5$ or that $f$ is continuous at a (for continuity, we also need $f(a)$ to be defined and equal to this limit).

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4. $\lim_{x \to \infty} \frac{1}{x}$ is equal to:

1. $\infty$

2. 1

3. 0

4. Does not exist

Show me the answer

Answer: 3. 0

Explanation:

• As x grows larger and larger (approaches infinity), the value of $\frac{1}{x}$ becomes smaller and smaller, approaching 0.

• Formally, $\lim_{x \to \infty} \frac{1}{x} = 0$.

• Similarly, $\lim_{x \to -\infty} \frac{1}{x} = 0$.

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Limit Laws and Indeterminate Forms

5. If $\lim_{x \to 2} f(x) = 3$ and $\lim_{x \to 2} g(x) = 4$, then $\lim_{x \to 2} [2f(x) - g(x)]$ equals:

1. 2

2. 5

3. 7

4. 14

Show me the answer

Answer: 1. 2

Explanation:

• Using limit laws (specifically, the difference law and constant multiple law): $\lim_{x \to 2} [2f(x) - g(x)] = 2 \cdot \lim_{x \to 2} f(x) - \lim_{x \to 2} g(x)$

• Substitute the given limits: $= 2 \cdot 3 - 4 = 6 - 4 = 2$

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6. Which of the following is an indeterminate form?

1. $\frac{0}{0}$

2. $\frac{\infty}{\infty}$

3. $0 \cdot \infty$

4. All of the above

Show me the answer

Answer: 4. All of the above

Explanation:

• Indeterminate forms are expressions that do not guarantee a specific limit value without further analysis. Common indeterminate forms include:

  • $\frac{0}{0}$ (e.g., $\lim_{x \to 0} \frac{\sin x}{x} = 1$, but $\lim_{x \to 0} \frac{x}{x^2} = \infty$)

  • $\frac{\infty}{\infty}$

  • $0 \cdot \infty$

  • $\infty - \infty$

  • $0^0$

  • $1^\infty$

  • $\infty^0$

• Techniques like factoring, rationalization, or L'Hôpital's Rule are used to evaluate them.

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7. $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ is equal to:

1. 0

2. 6

3. Does not exist

4. $\infty$

Show me the answer

Answer: 2. 6

Explanation:

• Direct substitution gives $\frac{0}{0}$, an indeterminate form.

• Factor the numerator: $x^2 - 9 = (x - 3)(x + 3)$

• Simplify: $\frac{(x - 3)(x + 3)}{x - 3} = x + 3$, for $x \neq 3$

• Now take the limit: $\lim_{x \to 3} (x + 3) = 3 + 3 = 6$

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Continuity

8. A function $f(x)$ is continuous at a point $x = a$ if:

1. $f(a)$ is defined

2. $\lim_{x \to a} f(x)$ exists

3. $\lim_{x \to a} f(x) = f(a)$

4. All of the above

Show me the answer

Answer: 4. All of the above

Explanation:

• The three conditions for continuity at a point $x = a$ are:

  1. $f(a)$ is defined (the point exists).

  2. The limit $\lim_{x \to a} f(x)$ exists.

  3. The limit equals the function value: $\lim_{x \to a} f(x) = f(a)$.

• If any one of these conditions fails, the function has a discontinuity at $x = a$.

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9. The function $f(x) = \frac{1}{x}$ is:

1. Continuous everywhere

2. Continuous for all $x \neq 0$

3. Continuous only at $x = 0$

4. Discontinuous everywhere

Show me the answer

Answer: 2. Continuous for all $x \neq 0$

Explanation:

• The function $f(x) = \frac{1}{x}$ is a rational function. Rational functions are continuous at every point in their domain.

• The domain of $f(x) = \frac{1}{x}$ is all real numbers except $x = 0$.

• Therefore, it is continuous for all $x \neq 0$.

• At $x = 0$, it is not defined, so it cannot be continuous there. It has an infinite discontinuity (vertical asymptote) at $x = 0$.

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10. The greatest integer function $f(x) = \lfloor x \rfloor$ (floor function) is discontinuous at:

1. All integer points

2. All rational points

3. All irrational points

4. It is continuous everywhere

Show me the answer

Answer: 1. All integer points

Explanation:

• The floor function $\lfloor x \rfloor$ gives the greatest integer less than or equal to x.

• At any integer value, say $x = n$:

  • The left-hand limit: $\lim_{x \to n^-} \lfloor x \rfloor = n - 1$

  • The right-hand limit: $\lim_{x \to n^+} \lfloor x \rfloor = n$

  • Since the left and right limits are not equal, the two-sided limit does not exist.

  • Therefore, the function has a jump discontinuity at every integer.

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Intermediate Value Theorem (IVT)

11. The Intermediate Value Theorem states that if a function $f$ is continuous on the closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$, then:

1. $f(c) = N$ for exactly one $c$ in $(a, b)$

2. $f(c) = N$ for at least one $c$ in $(a, b)$

3. $f(c) = 0$ for some $c$ in $(a, b)$

4. $f$ is differentiable on $(a, b)$

Show me the answer

Answer: 2. $f(c) = N$ for at least one $c$ in $(a, b)$

Explanation:

• The IVT guarantees the existence of at least one value $c$ in the open interval $(a, b)$ such that $f(c) = N$.

• It does not guarantee uniqueness. There could be multiple such $c$ values.

• A key application is proving that equations have roots. If $f(a)$ and $f(b)$ have opposite signs, then by taking $N = 0$, the theorem guarantees a root in $(a, b)$.

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12. Which condition is necessary to apply the Intermediate Value Theorem?

1. The function must be differentiable on $(a, b)$

2. The function must be continuous on $[a, b]$

3. The function must be one-to-one on $[a, b]$

4. $f(a)$ and $f(b)$ must have the same sign

Show me the answer

Answer: 2. The function must be continuous on $[a, b]$

Explanation:

• The key hypothesis of the IVT is that the function is continuous on the closed interval $[a, b]$.

• Differentiability is a stronger condition and is not required.

• The function does not need to be one-to-one.

• For the typical root-finding application, $f(a)$ and $f(b)$ must have opposite signs.

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Limits at Infinity and Asymptotes

13. If the degree of the numerator is less than the degree of the denominator in a rational function $R(x) = \frac{P(x)}{Q(x)}$, then $\lim_{x \to \pm \infty} R(x)$ is:

1. $\infty$

2. The ratio of the leading coefficients

3. 0

4. Does not exist

Show me the answer

Answer: 3. 0

Explanation:

• For limits at infinity of rational functions:

  • If degree(Numerator) < degree(Denominator): The limit is 0. The x-axis ($y=0$) is a horizontal asymptote.

  • If degree(Numerator) = degree(Denominator): The limit is the ratio of the leading coefficients.

  • If degree(Numerator) > degree(Denominator): The limit is $\infty$ or $-\infty$ (or DNE), and there is an oblique/slant asymptote.

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14. The line $x = a$ is a vertical asymptote of the function $y = f(x)$ if:

1. $\lim_{x \to a} f(x) = 0$

2. $\lim_{x \to a} f(x) = \infty$ or $-\infty$

3. $\lim_{x \to \infty} f(x) = a$

4. $f(a) = 0$

Show me the answer

Answer: 2. $\lim_{x \to a} f(x) = \infty$ or $-\infty$

Explanation:

• A vertical asymptote occurs at $x = a$ if the function's values increase or decrease without bound as x approaches a from the left, right, or both.

• Formally, at least one of these one-sided limits is infinite: $\lim_{x \to a^-} f(x) = \pm\infty$ or $\lim_{x \to a^+} f(x) = \pm\infty$.

• For rational functions, vertical asymptotes occur at the zeros of the denominator (provided they are not also zeros of the numerator).

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Squeeze Theorem

15. The Squeeze Theorem is useful for finding limits when:

1. The function is piecewise defined

2. The function can be bounded between two other functions whose limits are known and equal

3. Direct substitution yields an indeterminate form

4. The function is discontinuous

Show me the answer

Answer: 2. The function can be bounded between two other functions whose limits are known and equal

Explanation:

• The Squeeze Theorem (or Sandwich Theorem) states: If $g(x) \le f(x) \le h(x)$ for all x near a (except possibly at a), and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$.

• It is particularly helpful for limits involving trigonometric functions (like $\lim_{x \to 0} \frac{\sin x}{x}$) or oscillating functions multiplied by a decaying function.