The lim y - a-siny-a-2 -tan- y-2 a- is -A- a-2 -B- 2 a-C- a-D- -a-

The correct option is D aπlim_y→ a(siny a2(tanπ y2a)) is of the form 0×∞ ⇒lim_y→ a(siny a2π y2a) is of the form 00 Using L'Hospitals' rule, we get lim_y→ a(( ...

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Standard XII

Mathematics

Monotonically Increasing Functions

The lim y → a...

Question

The $lim y \to a [(sin \frac{y - a}{2}) \cdot (tan \frac{π y}{2 a})]$ is :

\frac{a}{2 π}

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\frac{2 a}{π}

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\frac{a}{π}

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\frac{- a}{π}

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Solution

The correct option is D $\frac{- a}{π}$
$lim y \to a (sin \frac{y - a}{2} (tan \frac{π y}{2 a}))$ is of the form $0 \times \infty$
$\Rightarrow lim y \to a ⎛ ⎜ ⎜ ⎜ ⎝ \frac{sin \frac{y - a}{2}}{cot \frac{π y}{2 a}} ⎞ ⎟ ⎟ ⎟ ⎠$ is of the form $\frac{0}{0}$

Using L'Hospitals' rule, we get
$lim y \to a ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{(cos \frac{y - a}{2}) \frac{1}{2}}{- {cosec}^{2} (\frac{π y}{2 a}) \frac{π}{2 a}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = \frac{- a}{π}$

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The limy→a[(siny−a2)⋅(tanπy2a)] is :

The correct option is D −aπlimy→a(siny−a2(tanπy2a)) is of the form 0×∞ ⇒limy→a⎛⎜ ⎜ ⎜⎝siny−a2cotπy2a⎞⎟ ⎟ ⎟⎠ is of the form 00 Using L'Hospitals' rule, we get limy→a⎛⎜ ⎜ ⎜ ⎜⎝(cosy−a2)12−cosec2(πy2a)π2a⎞⎟ ⎟ ⎟ ⎟⎠=−aπ

The $lim y \to a [(sin \frac{y - a}{2}) \cdot (tan \frac{π y}{2 a})]$ is :