The value of limt-0-1t2-2tt2sin t dt equals to

Let L=lim_t→0( 1t^2·∫_2t^t^2sin t dt ) (00) form Applying L'hospital rule, we get ⇒ L=lim_t→0sin t^2·2t sin 2t· 2 2t ⇒ L=lim_t→0( sin t^2+2sin 2t2t) ⇒ L=0+2=2 ...

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Standard XII

Mathematics

Function Defintion

The value of ...

Question

The value of $lim t \to 0 ⎛ ⎜ ⎜ ⎝ \frac{- 1}{t^{2}} \cdot t^{2} \int 2 t sin t d t ⎞ ⎟ ⎟ ⎠$ equals to

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Solution

Let $L = lim t \to 0 ⎛ ⎜ ⎜ ⎝ \frac{- 1}{t^{2}} \cdot t^{2} \int 2 t sin t d t ⎞ ⎟ ⎟ ⎠ (\frac{0}{0}) form$
Applying L'hospital rule, we get
$\Rightarrow L = lim t \to 0 \frac{sin t^{2} \cdot 2 t - sin 2 t \cdot 2}{- 2 t}$
$\Rightarrow L = lim t \to 0 (- sin t^{2} + \frac{2 sin 2 t}{2 t})$
$\Rightarrow L = 0 + 2 = 2$

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The value of limt→0⎛⎜ ⎜⎝−1t2⋅t2∫2tsintdt⎞⎟ ⎟⎠ equals to

Let L=limt→0⎛⎜ ⎜⎝−1t2⋅t2∫2tsintdt⎞⎟ ⎟⎠ (00) form Applying L'hospital rule, we get ⇒L=limt→0sint2⋅2t−sin2t⋅2−2t ⇒L=limt→0(−sint2+2sin2t2t) ⇒L=0+2=2

The value of $lim t \to 0 ⎛ ⎜ ⎜ ⎝ \frac{- 1}{t^{2}} \cdot t^{2} \int 2 t sin t d t ⎞ ⎟ ⎟ ⎠$ equals to