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Chapter 1: Problem 54

$\lim _{x \rightarrow 0} \frac{\ell n\left(1+x+x^{2}\right)+\ell n\left(1-x+x^{2}\right)}{\sec x-\cos x}$ is equal to (A) 1 (B) \(-1\) (C) 0 (D) \(\infty\)

Short Answer

Expert verified
In order to find the limit of the given expression as x approaches 0, we first checked if the expression resulted in an indeterminate form (\(\frac{0}{0}\)). Since it did, we applied L'H么pital's rule to simplify the expression and reevaluate the limit. Upon applying L'H么pital's rule twice, the final limit of the expression was found to be 0. Therefore, the correct answer is (C) 0.

Step by step solution

01

Check if the expression is an indeterminate form

First, let us evaluate the given expression as x approaches 0. Numerator: \(\lim_{x\to 0} \ln(1+x+x^2) = \ln(1+0+0) = \ln(1) = 0\) \(\lim_{x\to 0} \ln(1-x+x^2) = \ln(1-0+0) = \ln(1) = 0\) Denominator: \(\lim_{x\to 0} \sec(x) = \sec(0) = 1\) \(\lim_{x\to 0} \cos(x) = \cos(0) = 1\) So, the expression becomes: \(\lim_{x \rightarrow 0} \frac{0 + 0}{1 - 1} = \frac{0}{0}\) Thus, the expression is an indeterminate form. Let's apply L'H么pital's rule.
02

Apply L'H么pital's Rule

Applying L'H么pital's rule, we differentiate the numerator and the denominator separately with respect to x. Then, we find the limit of the new expression. Differentiate the Numerator: \(\frac{d}{dx}(\ln(1+x+x^2) + \ln(1-x+x^2)) = \frac{1 + 2x}{1+x+x^2} + \frac{-1 + 2x}{1-x+x^2}\) Differentiate the Denominator: \(\frac{d}{dx}(\sec x - \cos x) = \sec x \tan x + \sin x\) New expression: \(\lim_{x\to 0} \frac{\frac{1 + 2x}{1+x+x^2} + \frac{-1 + 2x}{1-x+x^2}}{\sec x \tan x + \sin x}\)
03

Evaluate the limit of the new expression

Now, let us evaluate the limit of the new expression as x approaches 0: Numerator: \(\lim_{x\to 0}\frac{1 + 2x}{1+x+x^2} = \frac{1}{1} = 1\) \(\lim_{x\to 0}\frac{-1 + 2x}{1-x+x^2} = \frac{-1}{1} = -1\) Denominator: \(\lim_{x\to 0}\sec x \tan x = \sec(0)\tan(0) = 1\cdot 0 = 0\) \(\lim_{x\to 0}\sin x = \sin(0) = 0\) So, the expression becomes: \(\lim_{x \rightarrow 0} \frac{1 -1}{0 +0} = \frac{0}{0}\) Since it is still an indeterminate form, let us apply L'H么pital's rule again.
04

Apply L'H么pital's Rule again

Differentiate the new numerator and the denominator, and then find the limit again: Differentiate the Numerator: \(\frac{d}{dx}(\frac{1 + 2x}{1+x+x^2} - \frac{-1 + 2x}{1-x+x^2}) = \frac{2}{(1+x+x^2)^2} - \frac{2}{(1-x+x^2)^2}\) Differentiate the Denominator: \(\frac{d}{dx}(\sec x \tan x + \sin x) = \sec x(\sec^2 x + \tan^2 x) + \cos x\) New expression: \(\lim_{x\to 0}\frac{\frac{2}{(1+x+x^2)^2} - \frac{2}{(1-x+x^2)^2}}{\sec x(\sec^2 x + \tan^2 x) + \cos x}\)
05

Evaluate the limit of the new expression

Now, let's again evaluate the limit as x approaches 0: Numerator: \(\lim_{x\to 0} \frac{2}{(1+x+x^2)^2} = \frac{2}{1} = 2\) \(\lim_{x\to 0} \frac{-2}{(1-x+x^2)^2} = \frac{-2}{1} = -2\) Denominator: \(\lim_{x\to 0} \sec x(\sec^2 x + \tan^2 x) = 1(1^2+0) = 1\) \(\lim_{x\to 0} \cos x = \cos(0) = 1\) So, the expression becomes: \(\lim_{x \rightarrow 0} \frac{2 -2}{1 +1} = \frac{0}{2}= 0\) Hence, the answer is (C) 0.

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Most popular questions from this chapter

\(\mathrm{A}_{0}\) 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

$\lim _{x \rightarrow \pi / 2} \frac{\sin x-(\sin x)^{\sin x}}{1-\sin x \ln \sin x}$ is equal to (A) 1 (B) zero (C) 2 (D) \(2 / 3\)

If $\mathrm{f}(\mathrm{x})=\left\\{\begin{array}{cl}\frac{\sin ([\mathrm{x}]+2 \mathrm{x})}{[\mathrm{x}]} & \text { if }[\mathrm{x}] \neq 0 \\ 0 & \text { if }[\mathrm{x}]=0\end{array}\right.\(, where \)[.]$ denotes the greatest integer function, then \(\lim _{x \rightarrow 0} f(x)\) is (A) 0 (B) 1 (C) \(-1\) (D) none of these

The true statement(s) is / are (A) If \(\lim _{x \rightarrow c} \mathrm{f}(\mathrm{x})=0\), then there must exist a number \(\mathrm{d}\) such that \(\mathrm{f}(\mathrm{d})<0.001\) (B) \(\lim _{x \rightarrow c} f(x)=L\), is equivalent to $\lim _{x \rightarrow c}(f(x)-L)=0$. (C) \(\lim _{x \rightarrow a}(f(x)+g(x))\) may exist even if the limits $\lim _{x \rightarrow i}$ \(\left(\mathrm{f}(\mathrm{x})\right.\) and $\lim _{\mathrm{x} \rightarrow \mathrm{a}}(\mathrm{g}(\mathrm{x})$ do not exist. (D) If \(\lim _{x \rightarrow a} f(x)\) exists and $\lim _{x \rightarrow a}(f(x)+g(x))$ does not exist, then \(\lim _{x \rightarrow a} g(x)\) does not exist.

\(\lim _{x \rightarrow 1} \frac{\tan (x-1) \cdot \log _{e} x^{x-1}}{|x-1|^{3}}\) is equal to (A) 1 (B) \(-1\) (C) 3 (D) None of these
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