← Back to 2. Limits and Continuity an Introduction

📝 Limits with radicals and square roots (14 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 14 questions available

←Previous Topic:Indeterminate form 0/0 limits📝 Limits with radicals and square rootsLimits of piecewise functions:Next Topic→

What is Limits with radicals and square roots?

Definition:
Limits involving radicals, such as square roots, often require rationalization to eliminate roots from the numerator or denominator, especially when direct substitution yields an indeterminate form like $0/0$ . Rationalization involves multiplying the numerator and denominator by the conjugate to simplify the expression, after which the limit can be evaluated using standard algebraic techniques and limit laws.

Example:
Evaluate $\lim_{x \to 4} \frac{\sqrt{x}-2}{x-4}$ .
Solution: $0/0$ ; multiply by conjugate: $\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)} = \frac{x-4}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x}+2}$ , so limit $= 1/(2+2)=1/4$ .

Reason:
This technique is essential for handling limits that arise in derivative calculations for root functions and in physics problems involving distances or rates, where radicals naturally appear and must be simplified.

Easy

Medium

Hard

📝 All Limits with radicals and square roots MCQs

Q1. What is the limit of $\frac{x-1}{\sqrt{x}-1}$ as $x \to 1$ ?

A.2 ✅

B.1

C.0

D.Does not exist

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Rationalizing the denominator gives $\frac{x-1}{\sqrt{x}-1}=\sqrt{x}+1$ . Substituting $x=1$ yields $\sqrt{1}+1=2$ . Hence the limit exists and equals 2, confirming the algebraic result.

Q2. Given that $\lim_{x\to 4} (\sqrt{x}-2)=0$ , which statement must be true about $\lim_{x\to 4}\frac{x-4}{\sqrt{x}-2}$ ?

A.0

B.2

C.does not exist

D.4 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Multiplying numerator and denominator by the conjugate $\sqrt{x}+2$ gives $\frac{x-4}{\sqrt{x}-2}=\sqrt{x}+2$ . As $x\to 4$ , $\sqrt{x}\to 2$ , so the expression tends to $2+2=4$ . Therefore the limit equals 4.

Q3. Compare the limits $\lim_{x\to 0} \frac{\sqrt{x+1}-1}{x}$ and $\lim_{x\to 0} \frac{x}{\sqrt{x+1}-1}$ . Which describes their relationship?

A.Both equal 0.5

B.First 0.5 and second 2 ✅

C.Both diverge

D.First 2 and second 0.5

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Using the derivative of $\sqrt{x+1}$ at 0 gives the first limit $\frac{1}{2\sqrt{1}}=0.5$ . The second limit is the reciprocal of the first, thus equals $2$ . Hence the correct relationship is option B.

Q4. Which algebraic manipulation correctly rationalizes the denominator of $\frac{1}{\sqrt{x}+3}$ ?

A.Multiply by (\sqrt{x}+3)/(\sqrt{x}+3)

B.Multiply numerator and denominator by (\sqrt{x}-3)

C.Multiply by (\sqrt{x}-3) only ✅

D.Multiply by (\sqrt{x}+3) only

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Rationalizing requires the conjugate $\sqrt{x}-3$ . Multiplying numerator and denominator by $\frac{\sqrt{x}-3}{\sqrt{x}-3}$ eliminates the radical in the denominator, yielding a rational expression. This is the standard technique for such fractions.

Q5. Suppose $f(x)=\frac{x-9}{\sqrt{x}-3}$ for $x\neq 9$ . Without simplifying, which limit value can be inferred from the continuity of $g(x)=\sqrt{x}$ at $x=9$ ?

A.3

B.6 ✅

C.0

D.Does not exist

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Since $\sqrt{x}$ is continuous at 9, $\sqrt{x}\to3$ as $x\to9$ . Rationalizing $f(x)$ gives $\sqrt{x}+3$ ; substituting the limit value yields $3+3=6$ . Thus the limit exists and equals 6.

Q6. Given the limits $L_1=\lim_{x\to a}\frac{x-a}{\sqrt{x}-\sqrt{a}}$ and $L_2=\lim_{x\to a}\frac{\sqrt{x}-\sqrt{a}}{x-a}$ , which relationship holds?

A.L1 = L2

B.L1 = 1/L2 ✅

C.Both 0

D.Both undefined

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Rationalizing $L_1$ gives $L_1=\sqrt{x}+\sqrt{a}\to2\sqrt{a}$ . $L_2$ is the reciprocal, approaching $1/(2\sqrt{a})$ . Therefore $L_1$ equals the reciprocal of $L_2$ , confirming option B.

Q7. When evaluating $\lim_{x\to 4} \frac{x-4}{\sqrt{x}-2}$ , why is it valid to cancel the factor $(\sqrt{x}-2)$ after rationalizing?

A.Because (\sqrt{x}-2) is never zero

B.Because limit excludes point where denominator zero ✅

C.Because (\sqrt{x}-2) is a constant

D.Because rationalization changes the function

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The cancellation occurs after multiplying by the conjugate, producing an equivalent expression for all $x\neq4$ . Since the limit considers values arbitrarily close to 4 but not equal to 4, the point where the denominator vanishes is excluded, making the cancellation legitimate.

Q8. If $\lim_{x\to 0}\frac{\sqrt{x+4}-2}{x}=k$ , which statement about $\lim_{x\to 0}\frac{x}{\sqrt{x+4}-2}$ is true?

A.It equals -k

B.It equals k^2

C.It does not exist

D.It equals 1/k ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The first limit evaluates to the derivative of $\sqrt{x+4}$ at 0, giving $k=1/4$ . The second limit is the reciprocal of the first, because the numerator and denominator are interchanged. Hence the limit equals $1/k$ , matching option D.

Q9. Which technique is most appropriate to evaluate $\lim_{x\to 0} \frac{\sqrt{x^2+1}-1}{x}$ ?

A.L'Hôpital's Rule

B.Series expansion

C.Multiply numerator and denominator by the conjugate ✅

D.Direct substitution

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The expression is an indeterminate $0/0$ . Multiplying by the conjugate $\frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}$ simplifies the numerator to $x^2$ , allowing cancellation of $x$ and straightforward limit evaluation. This is the most efficient method.

Q10. Given that $\lim_{x\to 1} (\sqrt{x}+1)=2$ , what does this imply about $\lim_{x\to 1}\frac{x-1}{\sqrt{x}-1}$ after rationalization?

A.It diverges

B.It equals 0

C.It equals 1

D.It equals 2 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Rationalizing the original fraction gives $\frac{x-1}{\sqrt{x}-1}=\sqrt{x}+1$ . Since the limit of $\sqrt{x}+1$ as $x\to1$ is 2, the original limit must also equal 2, confirming the result.

Q11. Which of the following limits is equivalent to $\lim_{x\to 4}\frac{x-4}{\sqrt{x}-2}$ after simplifying?

A.lim 1/(sqrt{x}+2)

B.lim (sqrt{x}+2) ✅

C.lim (sqrt{x}-2)

D.lim 1/(sqrt{x}-2)

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Rationalizing the fraction yields $\frac{x-4}{\sqrt{x}-2}=\sqrt{x}+2$ . Therefore the original limit equals $\lim_{x\to4}(\sqrt{x}+2)$ , which is option B.

Q12. In the limit process, why is it permissible to multiply by a factor of 1 such as $\frac{\sqrt{x}+1}{\sqrt{x}+1}$ ?

A.It removes radicals

B.It does not affect the limit because it's identically 1 ✅

C.It changes the function value

D.It only works for integer x

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Multiplying by $\frac{\sqrt{x}+1}{\sqrt{x}+1}=1$ leaves the expression unchanged for every $x$ in the domain. Since limits depend on the behavior of the function, not on algebraic form, the limit remains the same, making the operation valid.

Q13. If $\lim_{x\to a} f(x)=L\neq 0$ and $g(x)=\frac{1}{f(x)}$ , what is $\lim_{x\to a} g(x)$ ?

A.Does not exist

B.L

C.1/L ✅

D.0

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: When $f(x)$ approaches a non‑zero finite value $L$ , its reciprocal approaches the reciprocal of that value. Hence $\lim_{x\to a} \frac{1}{f(x)} = \frac{1}{L}$ . This follows directly from the continuity of the reciprocal function at non‑zero points.

Q14. Which statement correctly describes the behavior of $\frac{x-1}{\sqrt{x}-1}$ as $x$ approaches 1 from the left?

A.It oscillates

B.It is undefined for x<1

C.It approaches -2

D.It approaches 2 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Rationalizing the expression gives $\sqrt{x}+1$ , which is defined for all $x\ge0$ . As $x$ approaches 1 from either side, $\sqrt{x}$ approaches 1, so the expression tends to $1+1=2$ . Thus the left‑hand limit equals 2, matching option D.

🔗 Related Topics (MCQs)

📝 Tangent line problem and slope MCQs 📝 Area problem and limits MCQs 📝 Limits intuitive understanding MCQs 📝 Left hand and right hand limits MCQs 📝 One sided vs two sided limits MCQs

←Previous Topic:Indeterminate form 0/0 limits📝 Limits with radicals and square rootsLimits of piecewise functions:Next Topic→