/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Let $\mathrm{P}(\mathrm{x})=\mat... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 49

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

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

Step by step solution

01

Rewrite the expression

First, rewrite the expression inside the limit as follows: \(\lim _{x \rightarrow 0} \frac{\sqrt[100]{1+P(x)}-1}{x}\) = \(\lim _{x \rightarrow 0}\frac{(\frac{1}{100})(1+P(x))^{\frac{1}{100}-1}(P'(x))}{1}\)
02

Apply L'H么pital's Rule

We can now apply L'H么pital'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: \(\lim _{x \rightarrow 0}\frac{(\frac{1}{100})(-\frac{99}{100})(1+P(x))^{\frac{1}{100}-2}(P'(x)^2) + (\frac{1}{100})(1+P(x))^{\frac{1}{100}-1}(P''(x))}{0}\)
03

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) \(\frac{1}{100}\).

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Most popular questions from this chapter

Assertion $(\mathbf{A}): \lim _{\mathrm{x} \rightarrow 0^{\circ}}\left(\mathrm{x}^{\mathrm{x}^{*}}-\mathrm{x}^{\mathrm{x}}\right)=-1$ Reason \((\mathbf{R}): \lim _{x \rightarrow 0^{\prime}} x^{x}(x-1)=-1\)

The true statement(s) is / are (A) If \(\mathrm{f}(\mathrm{x})<\mathrm{g}(\mathrm{x})\) for all $\mathrm{x} \neq \mathrm{a}\(, then \)\lim _{\mathrm{x} \rightarrow \mathrm{a}} \mathrm{f}(\mathrm{x})<\lim _{x \rightarrow a} \mathrm{~g}(\mathrm{x})$. (B) If \(\lim _{x \rightarrow c} \mathrm{f}(x)=0\) and \(|g(x)| \leq M\) for a fixed number \(M\) and all \(x \neq c\), then \(\lim f(x) \cdot g(x)=0\) (C) If \(\lim _{x \rightarrow c} \mathrm{f}(\mathrm{x})=\mathrm{L}\), then $\lim _{\mathrm{x} \rightarrow \mathrm{c}}|\mathrm{f}(\mathrm{x})|=|\mathrm{L}|$ and conversely if \(\lim |\mathrm{f}(\mathrm{x})|=|\mathrm{L}|\) then $\lim _{x \rightarrow \infty} \mathrm{f}(\mathrm{x})=\mathrm{L}$. (D) If \(\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})\) for all real number other then \(\mathrm{x}=0\) and \(\lim _{x \rightarrow 0} f(x)=L\), then $\lim _{x \rightarrow 0} g(x)=L$

Let $\mathrm{f}(\mathrm{x})=\left[\begin{array}{ll}\mathrm{mx}^{2}+\mathrm{n} & \text { for } \quad \mathrm{x}<0 \\ \mathrm{n} x+\mathrm{m} & \text { for } 0 \leq \mathrm{x} \leq 1 \\ \mathrm{n} \mathrm{x}^{3}+\mathrm{m} & \text { for } \quad \mathrm{x}>1\end{array}\right.\( where \)\mathrm{m}, \mathrm{n} \in \mathrm{R}$ then which of the following must be correct (A) \(\lim _{x \rightarrow 0} f(x)\) exist for all values of \(m\) and \(n\). (B) \(\lim _{x \rightarrow 0} f(x)\) exists only if \(m=n\). (C) \(\lim _{x \rightarrow 0} f(x)\) exists for all values of \(m\) and \(n\). (D) \(\lim _{x \rightarrow 1} f(x)\) exists for no values of \(m\) and \(n\).

\(\mathrm{A}_{0}\) is an equilateral triangle of unit area, \(\mathrm{A}_{0}\) is divided into four equal parts, each an equilateral triangle, by joining the mid points of the sides of \(\mathrm{A}_{0}\). The central triangle is removed. Treating the remaining three triangles in the same way of division as was done to \(\mathrm{A}_{0}\), and this process is repeated \(\mathrm{n}\) times. The sum of the area of the triangles removed in \(\mathrm{S}_{\mathrm{n}}\) then $\lim _{\mathrm{n} \rightarrow \infty} \mathrm{S}_{\mathrm{n}}$ is (A) \(1 / 2\) (B) 1 (C) \(-1\) (D) 2

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