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Local Maxima

Which one of ...

Question

Which one of the following limit does not tends to unity?

lim t \to 0 \frac{sin (tan t)}{sin t}

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lim x \to \frac{π}{2} \frac{sin (cos x)}{cos x}

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lim x \to 0 \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x}

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lim x \to 0 \frac{\sqrt{x^{2}}}{x}

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Solution

The correct option is C $lim x \to 0 \frac{\sqrt{x^{2}}}{x}$
A) $lim t \to 0 \frac{sin (tan t)}{sin t}$
Applying L'hospital rule
$= lim t \to 0 \frac{cos (tan t) (1 + {(tan x)}^{2})}{cos t}$
$\frac{{lim}_{t \to 0} cos (tan t) \times {lim}_{t \to 0} (1 + {(tan x)}^{2})}{{lim}_{t \to 0} cos t} = \frac{1 \times 1}{1} = 1$

B) $lim x \to \frac{π}{2} \frac{sin (cos x)}{cos x}$
Applying L'hospital rule
$lim x \to \frac{π}{2} \frac{sin x (- cos (cos x))}{- sin x} = lim x \to \frac{π}{2} cos (cos x) = 1$

C) $lim x \to 0 \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x}$
Applying l'hospital rule
$= lim x \to 0 \frac{\frac{1}{2 \sqrt{1 + x}} + \frac{1}{2 \sqrt{1 - x}}}{1} = 1$

D) $lim x \to 0 \frac{\sqrt{x^{2}}}{x} = lim x \to 0 \frac{\pm x}{x} = \pm 1$

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