The value of _ x → 0 ^ - 4(2 + (3)/(x)) + 5 ( 2 ^ (1)/(x) ) 2(1 + (6)/(x)) + 6 ( 2 ^ (1)/(x) ) is equal to

Correct answer: A. (5)/(6) _ x → 0 ^ - (2(4 + (6)/(x)) + 5. 2(1)/(x))/(2(1 + (6)/(x)) + 6. 2(1)/(x)) = _ x → 0 ^ - 2(1)/(x)\ 16(2(5)/(x)) + 5\ 2 ^ (1)/(x) \ 2(2(5)/(x)) + 6\ _ x → 0 ^ - (16(2(5)/(x)) + 5)/(2(2(5)/(x)) + 6) = (16(0) + 5)/(2(0) + 6) = (5)/(6)

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Limits

The value of _ x → 0 ^ - 4(2 + (3)/(x)) + 5 ( 2 ^ (1)/(x) ) 2(1 +...

The value of

\lim_{x \to {0}^{-}}\dfrac{4\left(2 + \dfrac{3}{x}\right) + 5 ( {2}^{\dfrac{1}{x}} )}{2\left(1 + \dfrac{6}{x}\right) + 6 ( {2}^{\dfrac{1}{x}} )}

is equal to

Options

A $\dfrac{5}{6}$
B $8$
C $16$
D $\dfrac{5}{2}$

Correct answer

A. $\dfrac{5}{6}$

Step-by-step solution

\lim_{x \to {0}^{-}}\dfrac{2\left(4 + \dfrac{6}{x}\right) + 5. 2\dfrac{1}{x}}{2\left(1 + \dfrac{6}{x}\right) + 6. 2\dfrac{1}{x}} = \lim_{x \to {0}^{-}}\dfrac{2\dfrac{1}{x}\left\{16\left(2\dfrac{5}{x}\right) + 5\right\}}{{2}^{\dfrac{1}{x}}\left\{2\left(2\dfrac{5}{x}\right) + 6\right\}}

\lim_{x \to {0}^{-}}\dfrac{16\left(2\dfrac{5}{x}\right) + 5}{2\left(2\dfrac{5}{x}\right) + 6} = \dfrac{16\left(0\right) + 5}{2\left(0\right) + 6} = \dfrac{5}{6}

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