Question

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

Short answer

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'Hôpital's rule to simplify the expression and reevaluate the limit. Upon applying L'Hôpital'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

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'Hôpital's rule.

Apply L'Hôpital's Rule

Applying L'Hôpital'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: \(\lim_{x\to 0} \frac{\frac{1 + 2x}{1+x+x^2} + \frac{-1 + 2x}{1-x+x^2}}{\sec x \tan x + \sin x}\)

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'Hôpital's rule again.

Apply L'Hôpital'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: \(\lim_{x\to 0}\frac{\frac{2}{(1+x+x^2)^2} - \frac{2}{(1-x+x^2)^2}}{\sec x(\sec^2 x + \tan^2 x) + \cos x}\)

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.