Question

The value of \(\lim _{x\rightarrow \infty }\left(\sqrt[3]{x^{3}+6x^{2}+1}-x\right)\) is:

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(\infty\)

Hint:

Use a^3 - b^3 = (a - b)(a^2 + ab + b^2) with a the cube root and b = x.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Let \(a=\sqrt[3]{x^3+6x^2+1}\) and \(b=x\). Then \(a-b=\dfrac{a^3-b^3}{a^2+ab+b^2}\).

Step 2: Numerator
\(a^3-b^3=6x^2+1\).

Step 3: Divide by x squared
\(\dfrac{6+\frac{1}{x^2}}{\left(\frac{a}{x}\right)^2+\frac{a}{x}+1}\).

Step 4: Take the Limit
As \(x\rightarrow\infty\), \(\dfrac{a}{x}\rightarrow1\), so the limit is \(\dfrac{6}{1+1+1}=2\), option (C).

Final Answer:
The limit is 2. \[ \boxed{\text{(C) }2} \]