\(\underset { x\rightarrow 0 }{ lim } \frac { 2sinx-sin2x }{ { x }^{ 3 } } \)= \(\underset { x\rightarrow 0 }{ lim } \frac { 2sinx-2sinx\quad cosx }{ { x }^{ 3 } } \)
= \(\underset { x\rightarrow 0 }{ lim } \frac { 2sinx(1-cosx) }{ { x }^{ 3 } } =2.\underset { x\rightarrow 0 }{ lim } \frac { sinx }{ x } \underset { x\rightarrow 0 }{ lim } \frac { { 2sin }^{ 2 }\frac { x }{ 2 } }{ { x }^{ 2 } } \)  \(\left[ \therefore 1-cosx=2{ sin }^{ 2 }\frac { x }{ 2 } \right] \) 
= \(4(1).\underset { x\rightarrow 0 }{ lim } \frac { { sin }^{ 2 }\frac { x }{ 2 } }{ \left( \frac { x }{ 2 } \right) ^{ 2 } \times \ \ 4 } \left[ \therefore \underset { \phi \rightarrow 0 }{ lim } \quad \frac { sin\phi }{ \phi } =1 \right] \) [Multiplying and dividing by 4in the denominator]
= \(\frac { 4 }{ 4 } .\underset { x\rightarrow 0 }{ lim } \left( \frac { sin\frac { 2 }{ x } }{ \frac { 2 }{ x } } \right) \)
=1 x 1 = 1

\(\underset { x\rightarrow -3 }{ lim } \frac { { x }^{ 3 }+27 }{ { x }^{ 5 }+243 } =\underset { x\rightarrow -3 }{ lim } \frac { { x }^{ 3 }-\left( { -3 } \right) ^{ 3 } }{ { x }^{ 5 }-\left( { -3 } \right) ^{ 5 } } \)
Dividing the numerator and denominator by (x -3)
= \(\frac { \underset { x\rightarrow -3 }{ lim } \frac { { x }^{ 3 }-({ 3 })^{ 3 } }{ x-3 } }{ \frac { { x }^{ 5 }-\left( 3 \right) ^{ 5 } }{ x-3 } } =\frac { \underset { x\rightarrow -3 }{ lim } \frac { { x }^{ 3 }\left( { -3 } \right) ^{ 3 } }{ x-3 } }{ \underset { x\rightarrow -3 }{ lim } \frac { { x }^{ 5 }-\left( { -3 } \right) ^{ 5 } }{ x-3 } } \)
\(=\frac { 3\left( -3 \right) ^{ 3-1 } }{ 5\left( -3 \right) ^{ 5-1 } } \quad \left[ \therefore \underset { x\rightarrow a }{ lim } \frac { { x }^{ n }-{ a }^{ n } }{ x-a } ={ n }a^{ n-1 } \right] \)
= \(\frac { 3\left( -3 \right) ^{ 2 } }{ 5\left( -3 \right) ^{ 4 } } =\frac { 3(9) }{ 5(81) } =\frac { 3 }{ 5\left( 9 \right) } =\frac { 1 }{ 15 } \)

Given e^y(x+1)=1    ....(1)
Differentiating with respect to 'x' we get,
\({ e }^{ y }(1)+(x+1){ e }^{ y }\frac { dy }{ dx } =0\)   [product rule]
\(\Rightarrow { e }^{ y }+(1)\frac { dy }{ dx } =0\quad [using\quad (1)]\)
\(\Rightarrow \frac { dy }{ dx } =-{ e }^{ y }\)...(2)
Differentiating again with respect to 'x' we get,
\(\frac { d }{ dx } \left( \frac { dy }{ dx } \right) =\frac { d }{ dx } \left( -{ e }^{ y } \right) \)
\(\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-{ e }^{ y }.\frac { dy }{ dx } \)
\(=\left( \frac { dy }{ dx } \right) \left( \frac { dy }{ dx } \right) \) [using (2)]
\(={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
Hence proved.

Given x = 15(t - sint)
Differentiating with respect to 't' we get,
\(\frac{dy}{dx}\) = 15(1 - cos t) Also y = 18(1 - cos t)
\(\frac{dy}{dx}\) = 18(sin t)
Now  \(\frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } =\frac { 18\sin { t } }{ 15\left( 1-\cos { t } \right) } =\frac { 6\sin { t } }{ 5\left( 1-\cos { t } \right) } \)
\(=\frac { 6\times 2\sin { \frac { t }{ 2 } } \cos { \frac { t }{ 2 } } }{ 5\times 2\sin ^{ 2 }{ \frac { t }{ 2 } } } \)      \(\left[ \because sin2A=2sinAcosA\ \ and\ 1-cosA=2{ sin }^{ 2 }\frac { A }{ 2 } \right] \quad \quad \)
\(=\frac { 6 }{ 5 } \cot { \left( \frac { t }{ 2 } \right) } \)

Let y = sin x. sin 2x. sin 3x
Taking logarithms on both sides we get,
log y = log (sin x. sin 2x. sin 3x)
= log (sin x) + log (sin 2x) + log (sin 3x)    [\(\therefore\) log ab = log a + log b]
Differentiating with respect to 'x' we get,
\(\frac { 1 }{ y } \frac { dy }{ dx } =\frac { 1 }{ sin\quad x } .\frac { d }{ dx } (sin\quad x)+\frac { 1 }{ sin\quad 2x } .\frac { d }{ dx } (sin\quad 2x)+\frac { 1 }{ sin\quad 3x } .\frac { d }{ dx } (sin\quad 3x)\)
\(\frac { 1 }{ y } \frac { dy }{ dx } =\frac { cos\quad x }{ sin\quad x } +\frac { 2cos2x }{ sin\quad 2x } =3.\frac { cos\quad 3x }{ sin\quad 3x } \)
= cot x + 2 cot 2x + 3 cot 3x
\(\Rightarrow \frac { dy }{ dx } =y[cotx+2cot2x+3cot3x]\)
\(\Rightarrow \frac { dy }{ dx } \)= sin x sin 2x sin 3x [cot x + 2 cot 2x + 3 cot 3x]

Given sin² x + cos² y = 1.
Differentiating with respect to 'x' we get,
\(2\sin { x } .\frac { d }{ dx } \left( \sin { x } \right) +2\cos { y } .\frac { d }{ dx } \left( \cos { y } \right) =0\)    
\(\Rightarrow 2\sin { x } \cos { x } +2\cos { y } \left( -\sin { y } \right) \frac { dy }{ dx } =0\)
\(\Rightarrow \sin { 2x } -2\sin { y } \cos { y } \left( \frac { dy }{ dx } \right) =0\)
\(\Rightarrow \sin { 2x } -\sin { 2y } \left( \frac { dy }{ dx } \right) =0\)
\(\Rightarrow \sin { 2x } =\sin { 2y } \left( \frac { dy }{ dx } \right) \)
\(\Rightarrow \left( \frac { dy }{ dx } \right) =\frac { \sin { 2x } }{ \sin { 2y } } \)

\(y=\sin ^{ -1 }{ \left( \sqrt { \cos { x } } \right) } \)
Differentiating with respect to 'x' we have
\(\frac { dy }{ dx } =\frac { d }{ dx } .\sin ^{ -1 }{ { \left( \cos { x } \right) }^{ \frac { 1 }{ 2 } } } =\frac { 1 }{ \sqrt { 1-{ \left( \sqrt { \cos { x } } \right) }^{ 2 } } } .\frac { d }{ dx } { \left( \cos { x } \right) }^{ \frac { 1 }{ 2 } }\)
\(=\frac { 1 }{ \sqrt { 1-\cos { x } } } .\frac { 1 }{ 2\sqrt { 1-\cos { x } } } .\frac { d }{ dx } \left( \cos { x } \right) \)
\(=\frac { 1 }{ 2\sqrt { \cos { x } } .\sqrt { 1-\cos { x } } } \left( -\sin { x } \right) =\frac { -\sin { x } }{ 2\sqrt { \cos { x } } .\sqrt { 1-\cos { x } } } \)

Given f(x) = x² -x² -sin x + 5
\(\underset { x\rightarrow \pi }{ lim } f(x)=\underset { x\rightarrow \pi }{ lim } ({ x }^{ 2 }-sinx+5)\)
\(={ \pi }^{ 2 }-sin\quad \pi +5={ \pi }^{ 2 }-0+5+5=5{ \pi }^{ 2 }\)  \([\therefore sin \ \pi=0]\)
Also, f(\(\pi\)) = \({ \pi }^{ 2 }-sin\quad \pi +5=\pi \quad -0+5=\pi +5\)
ஃ  \(\underset { x\rightarrow \pi }{ lim } f(x)=f(x)\)
ஃ  f(x) is continuous at x = \(\pi\)

Given f(x) = [x].
Let a be any integer , then [a] = a
L[f(x)]_x=0 = \(\underset { x\rightarrow 0^{ - } }{ lim } f(x)=\underset { h\rightarrow 0^{ - } }{ lim } f(o-h)\)
\(=\underset { h\rightarrow 0^{ - } }{ lim } [a-h]=[a-o]=[a-1]\)  ....(1)
R[f(x)]_{x=0 =} \(\underset { x\rightarrow 0^{ + } }{ lim } f(x)=\underset { h\rightarrow 0 }{ lim } f(0+h)\)
\(=\underset { h\rightarrow 0 }{ lim } [a+h]=[a+0]=a\)   ...(2)
From (1) and (2), L[f(x)]_{x=0 \(\neq \)} R[f(x)]_x=0
So, f(x) is discontinuous at x =a where a is an arbitrary integral value
Thus, f(x) is discontinuous at all integral points

Given \(y=\sqrt { x } +\frac { 1 }{ \sqrt { x } } .\Rightarrow y=\frac { x+1 }{ \sqrt { x } } \)  ......(1)
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } =\frac { \sqrt { x } .\frac { d }{ dx } (x+1)-(x+1).\frac { d }{ dx } (\sqrt { x } ) }{ { (\sqrt { x } })^{ 2 } } \)
\(=\frac { \sqrt { x } (1)-(x+1).\frac { 1 }{ 2\sqrt { x } } }{ x } =\frac { \frac { 2x-(x+1) }{ 2\sqrt { x } } }{ x } \)
\(\frac { dy }{ dx } =\frac { 2x-x-1 }{ 2x.\sqrt { x } } =\frac { x-1 }{ 2x\sqrt { x } } \)...(2)
LHS=\(2x\left( \frac { dy }{ dx } \right) +y\)
\(=2x\left( \frac { x-1 }{ 2x\sqrt { x } } \right) +\frac { x+1 }{ \sqrt { x } }\)
 [From (1) and (2)]
\(=\frac { x-1 }{ \sqrt { x } } +\frac { x+1 }{ \sqrt { x } } =\frac { x-1+x+1 }{ \sqrt { x } } =\frac { 2x }{ \sqrt { x } } =\frac { 2\sqrt { x } .\sqrt { x } }{ \sqrt { x } } =2\sqrt { x } \) = RHS.
Hence proved