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📝 Limits with radicals at infinity (15 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 15 questions available

←Previous Topic:Rational functions limits at infinity📝 Limits with radicals at infinityEnd behavior and limits of trig functions:Next Topic→

What is Limits with radicals at infinity?

Definition:
Limits at infinity involving radicals, such as square roots, often require rationalization or factoring out the highest power of $x$ from the radicand to simplify. For expressions like $\sqrt{x^2 + ax}$ , the dominant term is $|x|$ , so careful handling of signs for $x \to \infty$ versus $x \to -\infty$ is needed, often using the conjugate or dividing by $x$ appropriately.

Example:
Evaluate $\lim_{x \to \infty} (\sqrt{x^2+1} - x)$ .
Solution: Multiply by conjugate: $\frac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{\sqrt{x^2+1}+x} = \frac{1}{\sqrt{x^2+1}+x} \to 0$ .

Reason:
This technique is important for handling limits in physics (e.g., relativistic energy) and engineering, where radicals appear in distance or error calculations, and understanding the asymptotic simplification is key to accurate approximations.

Easy

Medium

Hard

📝 All Limits with radicals at infinity MCQs

Q1. What technique should be applied to the expression $\sqrt{x^{6}+5}-x^{3}$ to prepare it for limit evaluation as $x\to +\infty$ ?

A.Differentiate the numerator

B.Multiply by its conjugate ✅

C.Apply L'Hôpital's rule

D.Introduce a new variable

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Multiplying by the conjugate $\bigl(\sqrt{x^{6}+5}+x^{3}\bigr)$ eliminates the radical in the numerator, producing a rational expression that can be simplified. This step is essential for handling the indeterminate form and revealing the dominant terms that determine the limit.

Q2. What is the value of $\displaystyle\lim_{x\to +\infty}\bigl(\sqrt{x^{6}+5}-x^{3}\bigr)$ ?

A.5

B.1

C.0 ✅

\infty

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: After rationalizing, the expression becomes $\frac{5}{\sqrt{x^{6}+5}+x^{3}}$ . As $x$ grows, the denominator behaves like $2x^{3}$ , so the whole fraction tends to $\frac{5}{2x^{3}} \to 0$ . Hence the limit equals zero.

Q3. Both $\sqrt{x^{6}+5x^{3}}$ and $x^{3}$ diverge to $+\infty$ as $x\to +\infty$ . What can be inferred about their difference $\sqrt{x^{6}+5x^{3}}-x^{3}$ ?

A.It approaches 0

B.It diverges to

\infty

C.It diverges to

-\infty

D.It approaches a finite positive number

\tfrac{5}{2}

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Rationalizing gives $\frac{5x^{3}}{\sqrt{x^{6}+5x^{3}}+x^{3}}$ . The denominator grows like $2x^{3}$ , so the fraction tends to $\frac{5}{2}$ . Thus the difference settles at the finite constant $5/2$ rather than vanishing or exploding.

Q4. Compare the asymptotic behavior of $f(x)=\sqrt{x^{6}+5}-x^{3}$ and $g(x)=\sqrt{x^{6}+5x^{3}}-x^{3}$ as $x\to +\infty$ . Which function attains the larger values for large $x$ ?

g(x)

is larger ✅

f(x)

is larger

C.Both approach the same limit

D.Cannot be determined from the given information

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The limit of $f(x)$ is 0, while $g(x)$ approaches $5/2$ . Since $5/2>0$ , for sufficiently large $x$ the values of $g(x)$ exceed those of $f(x)$ . Therefore $g(x)$ is the larger function asymptotically.

Q5. Which standard algebraic rule fails when manipulating expressions containing $+\infty$ or $-\infty$ ?

A.Commutative property

B.Associative property

C.Cancellation law ✅

D.Distributive law

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The cancellation law (e.g., subtracting the same infinite quantity from both sides) is not valid for $\pm\infty$ . While $\infty-\infty$ is an indeterminate form, one cannot simply cancel infinities as with finite numbers, leading to incorrect conclusions.

Q6. Given that $\displaystyle\lim_{x\to +\infty}\bigl(\sqrt{x^{6}+5}-x^{3}\bigr)=0$ , what does this tell about the growth rates of the two terms inside the limit?

\sqrt{x^{6}+5}

grows faster than

x^{3}

B.Both have the same leading term

x^{3}

✅

x^{3}

grows faster than the radical

D.Their growth rates are unrelated

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The radical expands to $x^{3}\sqrt{1+5/x^{6}} = x^{3}\bigl(1+ \tfrac{5}{2x^{6}}+o(1/x^{6})\bigr)$ . The leading term $x^{3}$ cancels with the subtracted $x^{3}$ , leaving a remainder that diminishes to zero, indicating identical dominant growth.

Q7. After rationalizing, which expression is equivalent to $\sqrt{x^{6}+5x^{3}}-x^{3}$ ?

\displaystyle\frac{5}{\sqrt{x^{6}+5x^{3}}+x^{3}}

✅

\displaystyle\frac{\sqrt{x^{6}+5x^{3}}-x^{3}}{5}

\displaystyle\frac{x^{3}}{5}

5\sqrt{x^{6}+5x^{3}}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Multiplying numerator and denominator by the conjugate yields $\frac{(x^{6}+5x^{3})-x^{6}}{\sqrt{x^{6}+5x^{3}}+x^{3}} = \frac{5x^{3}}{\sqrt{x^{6}+5x^{3}}+x^{3}}$ . Simplifying the factor $x^{3}$ gives the stated equivalent expression.

Q8. Which of the following steps is invalid when evaluating $\displaystyle\lim_{x\to +\infty}\bigl(\sqrt{x^{6}+5}-x^{3}\bigr)$ ?

A.Multiply by the conjugate

B.Factor out

x^{3}

from the radical

C.Apply the substitution

t=1/x

D.Cancel the

x^{3}

terms directly ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Directly canceling $x^{3}$ from $\sqrt{x^{6}+5}$ and the subtracted term assumes $\sqrt{x^{6}+5}=x^{3}$ , which is only approximately true for large $x$ . The correct approach requires rationalization or series expansion, not outright cancellation.

Q9. Define $h(x)=\frac{\sqrt{x^{6}+5}-x^{3}}{x^{2}}$ . What is $\displaystyle\lim_{x\to +\infty}h(x)$ ?

\tfrac{5}{2}

B.0 ✅

\infty

D.5

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Using the rationalized form $\frac{5}{\sqrt{x^{6}+5}+x^{3}}$ and dividing by $x^{2}$ gives $\frac{5}{x^{2}\bigl(\sqrt{x^{6}+5}+x^{3}\bigr)}\approx\frac{5}{2x^{5}}$ . As $x\to\infty$ , this expression tends to zero.

Q10. If the constant 5 in $\sqrt{x^{6}+5}-x^{3}$ is replaced by any positive constant $k$ , what is the limit as $x\to +\infty$ ?

A.0 ✅

\tfrac{k}{2}

\infty

D.Undefined

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The same rationalization yields $\frac{k}{\sqrt{x^{6}+k}+x^{3}}$ . The denominator behaves like $2x^{3}$ for large $x$ , so the fraction tends to $\frac{k}{2x^{3}}\to0$ regardless of the finite constant $k$ .

Q11. For large $x$ , what is the sign of $p(x)=\sqrt{x^{6}+5x^{3}}-x^{3}$ ?

A.Negative

B.Zero

C.Positive ✅

D.It alternates sign

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Since the numerator after rationalization is $5x^{3}$ and the denominator $\sqrt{x^{6}+5x^{3}}+x^{3}$ is always positive for $x>0$ , the entire expression is positive for all sufficiently large $x$ .

Q12. Which transformation most effectively simplifies the limit $\displaystyle\lim_{x\to +\infty}\bigl(\sqrt{x^{6}+5x^{3}}-x^{3}\bigr)$ ?

A.Apply L'Hôpital's rule

B.Rationalize the expression ✅

C.Use a power series expansion

D.Estimate the graph visually

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Rationalizing by multiplying with the conjugate turns the difference of radicals into a rational fraction, exposing the dominant terms and allowing direct evaluation of the limit without the need for differentiation or higher‑order series.

Q13. Evaluate $\displaystyle\lim_{x\to +\infty}\frac{\sqrt{x^{6}+5x^{3}}-x^{3}}{\sqrt{x^{6}+5}-x^{3}}$ .

A.0

\tfrac{5}{2}

\infty

D.Does not exist ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The numerator approaches $5/2$ while the denominator behaves like $\frac{5}{2x^{3}}$ . Their ratio is therefore approximately $\frac{5/2}{5/(2x^{3})}=x^{3}$ , which diverges to $\infty$ as $x\to\infty$ .

Q14. Using a series expansion, show that $\sqrt{x^{6}+5}-x^{3}= \frac{5}{2x^{3}}+O\!\left(\frac{1}{x^{9}}\right)$ . Which term dominates the difference for large $x$ ?

\frac{5}{2x^{3}}

✅

\frac{5}{x^{3}}

\frac{5}{x^{6}}

5x^{3}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Expanding $\sqrt{1+5/x^{6}}$ yields $1+\frac{5}{2x^{6}}-\frac{25}{8x^{12}}+\dots$ . Multiplying by $x^{3}$ gives $x^{3}+\frac{5}{2x^{3}}+O(1/x^{9})$ . Subtracting $x^{3}$ leaves the dominant term $5/(2x^{3})$ .

Q15. Find $L=\displaystyle\lim_{x\to +\infty}\Bigl[\bigl(\sqrt{x^{6}+5x^{3}}-x^{3}\bigr)-\bigl(\sqrt{x^{6}+5}-x^{3}\bigr)\Bigr]$ .

A.0

B.5

\tfrac{5}{2}

✅

D.Undefined

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The first bracket tends to $5/2$ while the second tends to 0, as shown earlier. Their difference therefore approaches $5/2-0=5/2$ . Hence the limit equals $5/2$ .

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←Previous Topic:Rational functions limits at infinity📝 Limits with radicals at infinityEnd behavior and limits of trig functions:Next Topic→