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$stack L t with x rightwards arrow 1 below open square brackets fraction numerator x to the power of 3 end exponent plus 2 x to the power of 2 end exponent plus x plus 1 over denominator x to the power of 2 end exponent plus 2 x plus 3 end fraction close square brackets to the power of fraction numerator 1 minus cos invisible function application left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis to the power of 2 end exponent end fraction end exponent$ is

e
e^1/2
1
$\sqrt{\frac{5}{6}}$

The correct answer is: $square root of fraction numerator 5 over denominator 6 end fraction end root$

$stack text Lt end text with x rightwards arrow 1 below open parentheses fraction numerator x to the power of 3 end exponent plus 2 x to the power of 2 end exponent plus x plus 1 over denominator x to the power of 2 end exponent plus 2 x plus 3 end fraction close parentheses to the power of fraction numerator 1 minus cos invisible function application left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis to the power of 2 end exponent end fraction end exponent equals stack text Lt end text with x rightwards arrow 1 below open parentheses fraction numerator 5 over denominator 6 end fraction close parentheses to the power of fraction numerator 1 minus cos invisible function application left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis to the power of 2 end exponent end fraction end exponent$
Put x − 1 = t
As x ® 1, t ® 0
given limit = $stack L t with t rightwards arrow 0 below open parentheses fraction numerator 5 over denominator 6 end fraction close parentheses to the power of fraction numerator 1 minus cos invisible function application t over denominator t to the power of 2 end exponent end fraction end exponent equals stack L t with t rightwards arrow 0 below open parentheses fraction numerator 5 over denominator 6 end fraction close parentheses to the power of fraction numerator 2 sin to the power of 2 end exponent invisible function application fraction numerator t over denominator 2 end fraction over denominator t to the power of 2 end exponent end fraction end exponent equals stack L t with t rightwards arrow 0 below open parentheses fraction numerator 5 over denominator 6 end fraction close parentheses to the power of open parentheses fraction numerator sin invisible function application t divided by 2 over denominator t divided by 2 end fraction close parentheses to the power of 2 end exponent. stack 2 with ̄ on top end exponent$
$equals open parentheses fraction numerator 5 over denominator 6 end fraction close parentheses to the power of 1 divided by 2 end exponent equals square root of fraction numerator 5 over denominator 6 end fraction end root$

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parallel

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If A and B are two independent events, then A and $stack B with ̄ on top$ are

If A and B are two independent events, then A and $stack B with ̄ on top$ are

parallel

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If $vertical line a cross times b vertical line equals 4$ and $vertical line a. b vertical line equals 2$ , then $vertical line a vertical line to the power of 2 end exponent vertical line b vertical line to the power of 2 end exponent equals$

If $vertical line a cross times b vertical line equals 4$ and $vertical line a. b vertical line equals 2$ , then $vertical line a vertical line to the power of 2 end exponent vertical line b vertical line to the power of 2 end exponent equals$

Let u = i + j, v = i – j and w = i + 2j + 3k. If n is a unit vector such that u.n = 0 and v.n = 0 then $vertical line w. n vertical line$ is equal to

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parallel

Angle between the line $r equals left parenthesis i plus 2 j minus k right parenthesis plus lambda rightwards arrow over short leftwards arrow left parenthesis i minus j plus k right parenthesis$ and the normal to the plane $r. left parenthesis 2 i minus j plus k right parenthesis equals 4$ is

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parallel

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parallel

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