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

Correct answer: C. 2 Let, sec x = t _ t → 1 t - t^ t 1 - t + (t) = _ t → 1 1 - t^ t ( 1 + ln t ) - 1 + (1)/(t) by L-hospital rule = _ t → 1 - t^ t ((1)/(t)) - t^ t (1 + ln t)^ 2 - 1 / t^ 2 again by L-hospital rule = (- 1 - 1)/(- 1) = 2

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Limits

The value of _ x → 0 sec x - ( sec x) ^ x 1 - sec x + ln (sec x) is...

The value of

\lim_{x \to 0}\dfrac{sec x - {\left(\text{ sec } x\right)}^{\text{ \sec x }}}

{1 - sec x + ln

\left(sec x\right)}

is equal to

Options

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

Correct answer

C. $2$

Step-by-step solution

Let,

sec x = t

\lim_{t \to 1}\dfrac{t - t^{t}}{1 - t + \ln \left(t\right)} = \lim_{t \to 1}\dfrac{1 - t^{t} ( 1 + ln t )}{- 1 + \dfrac{1}{t}}

{by L-hospital rule}

= \lim_{t \to 1}\dfrac{- t^{t}\left(\dfrac{1}{t}\right) - t^{t}\left(1 + ln t\right)^{2}}{- 1 / t^{2}}

{again by L-hospital rule}

= \dfrac{- 1 - 1}{- 1} = 2

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