Question

If $y = \left[(x+1)(2x+1)(3x+1)\cdots (nx+1)\right]^n$, then $\left.\frac{dy}{dx}\right|_{x=0}$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{4\sqrt{2}}$
• C. $\frac{-1}{4\sqrt{2}}$
• D. $\frac{-1}{\sqrt{2}}$

Hint:

Calculus Tip: When dealing with limits at infinity involving nested radicals, double rationalization is a common and effective technique to eliminate the indeterminate forms before factoring out the highest power of $x$.

Answer 1

The Correct Option is B

Concept: Calculus - Limits at Infinity and Rationalization.

Step 1: Rationalize the main expression. The given limit is of the form $\infty \times 0$. To resolve this, multiply and divide the expression within the brackets by its conjugate: $\sqrt{x^{2}+\sqrt{1+x^{4}}} + x\sqrt{2}$. The numerator becomes $(x^{2}+\sqrt{1+x^{4}}) - (x\sqrt{2})^2 = x^{2}+\sqrt{1+x^{4}} - 2x^{2} = \sqrt{1+x^{4}} - x^{2}$. The expression is now: $\lim_{x\rightarrow\infty} \frac{x^{3}(\sqrt{1+x^{4}}-x^{2})}{\sqrt{x^{2}+\sqrt{1+x^{4}}}+x\sqrt{2}}$.

Step 2: Rationalize the new numerator. The numerator still leads to an indeterminate form. Rationalize it again by multiplying the numerator and denominator by its conjugate, $\sqrt{1+x^{4}} + x^{2}$. The new numerator becomes $(1+x^{4}) - (x^{2})^2 = 1+x^{4}-x^{4} = 1$. The limit expression is now: $\lim_{x\rightarrow\infty} \frac{x^{3}(1)}{(\sqrt{x^{2}+\sqrt{1+x^{4}}}+x\sqrt{2})(\sqrt{1+x^{4}}+x^{2})}$.

Step 3: Factor out the highest powers of $x$ from the denominator. Factor out $x$ from the first bracket and $x^2$ from the second bracket to match the $x^3$ in the numerator. First bracket: extract $x = \sqrt{x^2}$ to get $x\left(\sqrt{1+\sqrt{\frac{1}{x^4}+1}} + \sqrt{2}\right)$. Second bracket: extract $x^2 = \sqrt{x^4}$ to get $x^2\left(\sqrt{\frac{1}{x^4}+1} + 1\right)$. Multiply these extracted factors: $x \cdot x^2 = x^3$.

Step 4: Cancel $x^3$ and prepare to evaluate the limit. Substitute the factored denominator back into the limit: $\lim_{x\rightarrow\infty} \frac{x^{3}}{x^{3}\left(\sqrt{1+\sqrt{\frac{1}{x^{4}}+1}}+\sqrt{2}\right)\left(\sqrt{\frac{1}{x^{4}}+1}+1\right)}$. The $x^3$ terms cancel out completely, leaving: $\lim_{x\rightarrow\infty} \frac{1}{\left(\sqrt{1+\sqrt{\frac{1}{x^{4}}+1}}+\sqrt{2}\right)\left(\sqrt{\frac{1}{x^{4}}+1}+1\right)}$.

Step 5: Evaluate the limit as $x \rightarrow \infty$. As $x$ approaches infinity, $\frac{1}{x^4}$ approaches 0. Substitute 0 for all $\frac{1}{x^4}$ terms: $\frac{1}{\left(\sqrt{1+\sqrt{0+1}}+\sqrt{2}\right)\left(\sqrt{0+1}+1\right)} = \frac{1}{\left(\sqrt{1+1}+\sqrt{2}\right)(1+1)} = \frac{1}{(\sqrt{2}+\sqrt{2})(2)} = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$. $$ \therefore \text{The value of the limit is } \frac{1}{4\sqrt{2}}. $$