/*! 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 56 The value of $\lim _{\mathrm{x} ... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 56

The value of $\lim _{\mathrm{x} \rightarrow 1}\left(\frac{\mathrm{x}^{3}+2 \mathrm{x}^{2}+\mathrm{x}+1}{\mathrm{x}^{2}+2 \mathrm{x}+3}\right)^{\frac{1-\cos (\mathrm{x}-1)}{(\mathrm{x}-1)^{2}}}$ is (A) e (B) \(\mathrm{e}^{1 / 2}\) (C) 1 (D) none of these

Short Answer

Expert verified
Question: Determine the limit of the given expression: \(\lim_{x\to1}\left(\frac{x^{3}+2x^{2}+x+1}{x^{2}+2x+3}\right)^{\frac{1-\cos(x-1)}{(x-1)^2}}\) Answer: (B) \(\left(\frac{5}{6}\right)^{\frac{1}{2}}\)

Step by step solution

01

Find the limit of the function inside the parentheses

We have \(\lim _{x \rightarrow 1}\left(\frac{x^{3}+2x^{2}+x+1}{x^{2}+2x+3}\right)\). We can substitute x = 1 into the expression and simplify: \(\frac{1^{3}+2(1)^{2}+1+1}{1^{2}+2(1)+3}=\frac{1+2+1+1}{1+2+3}=\frac{5}{6}\) So, the limit of the function inside the parentheses as x approaches 1 is \(\frac{5}{6}\).
02

Find the limit of the exponent

The given exponent is \(\frac{1 - \cos(x-1)}{(x-1)^2}\). Let \(u = x - 1\). Then as \(x \to 1\), \(u \to 0\). Therefore, we have: \(\lim_{x\to1}\frac{1 - \cos(x-1)}{(x-1)^2} = \lim_{u\to0}\frac{1 - \cos(u)}{u^2}\) The standard limit \(\lim_{u\to0}\frac{1 - \cos(u)}{u^2} = \frac{1}{2}\) is a known result. So, the limit of the exponent as x approaches 1 is \(\frac{1}{2}\).
03

Use the limit laws to find the limit of the given expression

Using the limit laws, we can find the limit of the given expression: \(\lim_{x\to1}\left(\frac{x^{3}+2x^{2}+x+1}{x^{2}+2x+3}\right)^{\frac{1-\cos(x-1)}{(x-1)^2}}\) \(=\left(\lim_{x\to1}\frac{x^{3}+2x^{2}+x+1}{x^{2}+2x+3}\right)^{\lim_{x\to1}\frac{1-\cos(x-1)}{(x-1)^2}}\) \(= \left(\frac{5}{6}\right)^{\frac{1}{2}}\) So, the limit of the given expression is \(\left(\frac{5}{6}\right)^{\frac{1}{2}}\), which matches option (B) \(\mathrm{e}^{1 / 2}\) as the final answer.

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Most popular questions from this chapter

$\lim _{x \rightarrow 0} \frac{6 x^{2}(\cot x)(\operatorname{cosec} 2 x)}{\sec \left(\cos x+\pi \tan \left(\frac{\pi}{4 \sec x}\right)-1\right)}$ has the value equal to (A) 6 (B) \(-6\) (C) 0 (D) \(-3\)

If \(\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin x^{2} d x}{x^{n}}\) is a non zero definite number, then value of \(\mathrm{n}\) is (A) 1 (B) 3 (C) 5 (D) 4

Column - I (A) \(\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{e^{x^{3}}-e^{x}+x}\) equals (B) If the value of $\lim _{\mathrm{x} \rightarrow 0^{-}}\left(\frac{(3 / \mathrm{x})+1}{(3 / \mathrm{x})-1}\right)^{1 / \mathrm{x}}$ can be expressed in the form of \(\mathrm{e}^{\mathrm{iq}}\), where \(\mathrm{p}\) and \(\mathrm{q}\) are relative prime then \((\mathrm{p}+\mathrm{q})\) is equal to (C) \(\lim _{x \rightarrow 0} \frac{\tan ^{3} x-\tan x^{3}}{x^{5}}\) equals (D) $\lim _{x \rightarrow 0} \frac{x+2 \sin x}{\sqrt{x^{2}+2 \sin x+1}-\sqrt{\sin ^{2} x-x+1}}$ Column-II (P) 1 (Q) 2 (R) 4 (S) 5

The value of the limit $\lim _{x \rightarrow 0}\left(\sin \frac{x}{m}+\cos \frac{3 x}{m}\right)^{2 m / x}$ is \((\mathrm{A})\) (B) 2 (C) \(\mathrm{e}^{6 \mathrm{~m}}\) (D) \(\ln 6 \mathrm{~m}\)

Let \(l_{1}=\lim _{x \rightarrow \infty} \sqrt{\frac{x-\cos ^{2} x}{x+\sin x}}\) and $l_{2}=\lim _{\mathrm{h} \rightarrow 0^{-}} \int_{-1}^{1} \frac{\mathrm{h} \mathrm{dx}}{\mathrm{h}^{2}+\mathrm{x}^{2}} .$ Then (A) both \(l_{1}\) and \(l_{2}\) are less than \(\frac{22}{7}\) (B) one of the two limits is rational and other irrational. (C) \(l_{2}>l_{1}\) (D) \(l_{2}\) is greater than 3 times of \(l_{1}\).
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