Question

Let $\mathrm{P}(\mathrm{x})=\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrm{x}^{2}+\mathrm{a}_{3} \mathrm{x}^{3}+\ldots \ldots .+\mathrm{a}_{100} \mathrm{x}^{100}\(, where \)\mathrm{a}_{1}=$ 1 and \(a_{i} \in R \forall i=2,3,4, \ldots, 100\) then \(\lim _{x \rightarrow 0} \frac{\sqrt[100]{1+P(x)}-1}{x}\) has the value equal to (A) 100 (B) \(\frac{1}{100}\) (C) 1 (D) 5050

Short answer

Answer: (B) \(\frac{1}{100}\)

Step-by-step solution

Rewrite the expression

First, rewrite the expression inside the limit as follows: \(\lim _{x \rightarrow 0} \frac{\sqrt[100]{1+P(x)}-1}{x}\) = \(\lim _{x \rightarrow 0}\frac{(\frac{1}{100})(1+P(x))^{\frac{1}{100}-1}(P'(x))}{1}\)

Apply L'Hôpital's Rule

We can now apply L'Hôpital's Rule to the expression. The rule states that if the expression is of the form "0/0" or "infinity/infinity," we can differentiate the numerator and denominator separately and take the limit of the new fraction. In this case, differentiating the numerator gives: \(\frac{d}{dx}((\frac{1}{100})(1+P(x))^{\frac{1}{100}-1}(P'(x))) = (\frac{1}{100})(-\frac{99}{100})(1+P(x))^{\frac{1}{100}-2}(P'(x)^2) + (\frac{1}{100})(1+P(x))^{\frac{1}{100}-1}(P''(x))\) Differentiating the denominator gives: \(\frac{d}{dx}(1) = 0\) So the limit becomes: \(\lim _{x \rightarrow 0}\frac{(\frac{1}{100})(-\frac{99}{100})(1+P(x))^{\frac{1}{100}-2}(P'(x)^2) + (\frac{1}{100})(1+P(x))^{\frac{1}{100}-1}(P''(x))}{0}\)

Apply properties of limits

Since the denominator of the fraction is now 0 (as a result of the differentiation), we can apply the properties of limits to solve this problem: \(\lim _{x \rightarrow 0}\frac{(\frac{1}{100})(-\frac{99}{100})(1+P(x))^{\frac{1}{100}-2}(P'(x)^2) + (\frac{1}{100})(1+P(x))^{\frac{1}{100}-1}(P''(x))}{0}\) = \(\lim _{x \rightarrow 0}(\frac{1}{100})(-\frac{99}{100})(1+P(x))^{\frac{1}{100}-2}(P'(x)^2) + \lim _{x \rightarrow 0} (\frac{1}{100})(1+P(x))^{\frac{1}{100}-1}(P''(x))\) Since \(\lim _{x \rightarrow 0} (1+P(x)) = 1+1 = 2\) and \(\lim _{x \rightarrow 0} (P'(x))\) and \(\lim _{x \rightarrow 0} (P''(x))\) are finite, we can evaluate the limits as follows: \(\lim _{x \rightarrow 0}(\frac{1}{100})(-\frac{99}{100})(1+P(x))^{\frac{1}{100}-2}(P'(x)^2) = 0\) \(\lim _{x \rightarrow 0} (\frac{1}{100})(1+P(x))^{\frac{1}{100}-1}(P''(x)) = (\frac{1}{100})(2^{\frac{1}{100}-1})(P''(0)) = \frac{1}{100}\) So the answer is: (B) \(\frac{1}{100}\).