The value of _ x → 0 ( 1 + x + x^ 2 ) + ( 1 - x + x^ 2 ) x - x & is equal to

Correct answer: B. 1 _ x → 0 (1 + x + x^ 2 ) + (1 - x + x^ 2 ) sec x - cos x = _ x → 0 [ (1 + x^ 2 ) ^ 2 - x^ 2 ] (1 - ^ 2 x) / cos x = _ x → 0 (1 + x^ 2 + x^ 4 ) sin x tan x = _ x → 0 (1 + x^ 2 (1 + x^ 2 )) x^ 2 (1 + x^ 2 ) · x^ 2 (1 + x^ 2 ) · 1 (sin x)/(x) · (tan x)/(x) · x^ 2 = 1 (as _ x → 0 (1 + x) x =1)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Limits

NTA Abhyas 2020 · Mathematics · Limits

The value of _ x → 0 ( 1 + x + x^ 2 ) + ( 1 - x + x^ 2 ) x - x & is...

The value of

\lim_{x \to 0}\dfrac{\log( 1 + x + x^{2} ) + \log( 1 - x + x^{2} )}{\sec x - \cos x}

& is equal to

Options

A $- 1$
B $1$
C $0$
D $2$

Correct answer

B. $1$

Step-by-step solution

\lim_{x \to 0} \dfrac{\log \left(1 + x + x^{2}\right) + \log \left(1 - x + x^{2}\right)}{sec x - cos x}

= \lim_{x \to 0} \dfrac{\log \left[{\left(1 + x^{2}\right)}^{2} - x^{2}\right]}{\left(1 - \cos^{2 } x\right) / cos x}

= \lim_{x \to 0} \dfrac{\log \left(1 + x^{2} + x^{4}\right)}{sin x tan x}

= \lim_{x \to 0} \dfrac{\log \left(1 + x^{2} \left(1 + x^{2}\right)\right)}{x^{2} \left(1 + x^{2}\right)} \cdot x^{2} \left(1 + x^{2}\right) \cdot \dfrac{1}{\dfrac{sin x}{x} \cdot \dfrac{tan x}{x} \cdot x^{2}}

= 1 \left(as \lim_{x \to 0} \dfrac{\log \left(1 + x\right)}

{x} =1\right)

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