Let the price of low-grade steel be Rs. p per tonne. Then the price of high-grade steel is Rs. 2p per tonne.
The total receipt per day is given by \(R=px+py=px+2p\left( \frac { 40-5x }{ 10-x } \right) \). Hence the problem is to maximise R . Now, simplifying and differentiating R with respect to x , we get
\(R=p\left( \frac { 80-{ x }^{ 2 } }{ 10-x } \right) \)
\(\frac { dR }{ dx } =p\left( \frac { { x }^{ 2 }-20x+80 }{ (10-x)^{ 2 } } \right) \)
\(\frac { dR }{ dx } =-\frac { 40P }{ \left( 10-x \right) ^{ 3 } } \)
Now, \(\frac { dR }{ dx } =0\Rightarrow { x }^{ 2 }-20x+80=0\) and hence \(x=10\pm 2\sqrt { 5 } \) 
At \(x=10-2\sqrt { 5 } ,\frac { { d }^{ 2 }R }{ { dx }^{ 2 } } <0\) and hence R will be maximum. If x \(x=10-2\sqrt { 5 } \) then \(y=5-\sqrt { 5 } \)  
Therefore the steel plant must produce low-grade and high-grade steels respectively in tonnes per day are \(10-2\sqrt { 5 } \)  and \(5-5\sqrt { 5 } \)

This is an indeterminate of the from 1∞
Let g(x) = (sin x)^{tan x}
Taking logarithm, we get
log (g(x)) = log (sin x)tan x
= tan x log (sin x)
Now, \(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } log(g(x))\) = \(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } \frac { log(sinx) }{ cotx } =\left( \frac { 0 }{ 0 } form \right) \)
= \(\frac { \underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } \frac { 1 }{ sinx } \times cosx }{ -{ cosec }^{ 2 }x } \) [By L' Hôpital Rule]
= \(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } -\frac { cosx }{ sinx\frac { 1 }{ { sin }^{ 2 }x } } \)
= \(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } cos \ x \ sin \ x=0\)
But \(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } log(g(x))=log(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } g(x))\)
\(\therefore log\left( \underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } g(x) \right) =log \ 0=1\)

\(\Rightarrow { e }^{ log }\left( \underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } g(x) \right) ={ e }^{ 1 }\)
\(\Rightarrow \underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } g(x)=e\Rightarrow \underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } { (sinx) }^{ tanx }=e^o=1\)

Let f(x) = tan^-1(x)
⇒ f(0) = tan-1(0) = 0
\({ f }^{ -1 }(x)=\frac { 1 }{ 1+{ x }^{ 2 } } ={ (1+{ x }^{ 2 }) }^{ -1 }\)
⇒ f^I(0) = 1
f^II(x) = -1(1 + x)^-2 (2x)
= -2x(1 + x²)^-2
⇒ f^II(0) = 0
= -2[-4x²(1 + x²)^-3 + (1 + x²)^-2]
f^III(x) = -2[x(-2)(1+x²)^-3 (2x) + (1 + x²)^-2]
⇒ f^III(0) = 2
f^IV(x) = -2[-4x²(-3) (1 + x²)^-4 (2x) + (1 + x²)-3(-8x) + (1 + x²)^-3(-2)(2x)]
⇒ f^IIl(0) = 2
= - 2 [24x³ (1 + x²)^-4-8x (1+ x²)³ -4x (1 + x²)³]
= - 2 [24x³ (1 + x²)^-4-12x (1+ x²)³]
f^IV(x) = 24[2x³ (1 + x²)-4 - x (1 + x²)^-3]
⇒ f^Il(0) = 0
f^V(x) = - 24 [2x³ (-4)(1 +x²)^-5(2x) + (1+ x²)^-4(6x² -x (- 3)(1 + x²)-4(2x) - (1 + x²)³]
= - 24 [16x⁴ (1 + x²)^-5+ 6x² (1+ x²)-4- (1 + x²)³]
= - 24 [16x⁴ (1 + x²)^-5+12x² (1+ x²)^-4- (1 + x²)³] 
⇒ fV(0) = 24
⇒ f(0) = tan-1 (0) = 0
⇒ f^I(0) = 1
⇒ f^II(0) = 0
⇒ f^III(0) = -2
⇒ f^IV(0) = 0
⇒ f^V(0) = 24
Maclaurin's series
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\frac { { f }^{ IV }(0) }{ 4! } { x }^{ 4 }+..\)
= \(0+\frac { 1 }{ 1! } x+0\frac { 2 }{ 3! } { x }^{ 3 }+0+\frac { 24 }{ 5! } { x }^{ 5 }+\)...
= x - \(\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 5 } }{ 5 } +\).......

Given AC = b and CD = a
Let hand r be the height and radius of the cylinder
ஃ AB = AC - BC = b - h
In the diagram

ΔABE - ΔACD
\(\therefore \frac { AB }{ BE } =\frac { AC }{ CD } \) 
\(\Rightarrow \frac { b-h }{ r } =\frac { b }{ a } \)
 \(\Rightarrow r=\frac { a\left( b-h \right) }{ b } \)
Volume of the cylinder = \({ \pi r }^{ 2 }h\)
= \(\pi \left( \frac { a }{ b } \left( b-b \right) ^{ 2 }h \right) \) 
= \(\pi \frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( b-h \right) ^{ 2 }h\)
= \(\pi \frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( { b }^{ 2 }+{ h }^{ 2 }-2bh \right) h\)
 \(V=\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( { b }^{ 2 }{ h }^{ 2 }+{ h }^{ 3 } \right) \)
Differentiating with respect to 'h',
\(\frac { dV }{ dh } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( { b }^{ 2 }+{ 3h }^{ 2 }-4bh \right) \)
\(\frac { dV }{ dh } =0\Rightarrow { b }^{ 2 }-4bh+{ 3j }^{ 2 }=0\)
\(\Rightarrow (b-3h)+(b-h)=0\Rightarrow b-3h=0\quad \)or b = h
\(\Rightarrow h=\frac { b }{ 3 } asb\neq h\) 
\(\frac { dV }{ { dh }^{ 2 } } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( 6\left( \frac { b }{ 3 } \right) -4b \right) \)
\(\left( \frac { { d }^{ 2 }V }{ { dh }^{ 2 } } \right) _{ at\ h=\frac { b }{ 3 } }=\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( 6\left( \frac { b }{ 3 } \right) -4b \right) \)
= \(\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi (-2b)<0\) 
ஃ V is maximum at \(h=\frac { b }{ 3 } \)
ஃ Maximum volume of the cylinder
\(V=\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( b-h \right) ^{ 2 }h\) 
= \(\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( b-\frac { b }{ 3 } \right) ^{ 2 }\left( \frac { b }{ 3 } \right) \left[ \therefore h=\frac { b }{ 3 } \right] \)
= \(\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( \frac { 2b }{ 3 } \right) ^{ 2 }\left( \frac { b }{ 3 } \right) \)
\(=\frac{\pi a^{2}}{\not b^{2}}\left(\frac{4\not b^{2}}{9}\right)\left(\frac{b}{3}\right)=\frac{4}{9}\left(\frac{1}{3} \pi a^{2} b\right)\)
= \(\frac { 4 }{ 9 } \)(Volume of the cone)
Hence, volume of the largest cylinder that can be hidden underneath is \(\frac { 4 }{ 9 } \) times volume of the cone.

Given \(f(x)=\frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } \)
\(\quad \underset { x\rightarrow 1^{ + } }{ lim } \frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } =\underset { x\rightarrow 1^{ + } }{ lim } \frac { { x }^{ 2 } }{ \left( x+1 \right) \left( x-1 \right) } \) 
= \(\underset { h\rightarrow { 0 }^{ + } }{ lim } \frac { \left( 1+h \right) ^{ 2 } }{ \left( 1+h+1 \right) \left( 1+h-1 \right) } \)
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { (1+h)^{ 2 } }{ (2+h)(h) } \infty \) 
Also  \(\underset { x\rightarrow 1^{ - } }{ lim } \frac { { x }^{ 2 } }{ (x+1)/(x-1) } =\underset { h\rightarrow 0^{ + } }{ lim } \frac { \left( 1-h \right) ^{ 2 } }{ \left( 1-h+1 \right) \left( 1-h-1 \right) } \)
= \(\underset { h\rightarrow { 0 }^{ + } }{ lim } \frac { \left( 1-h \right) ^{ 2 } }{ (2-h)(-h) } =-\infty \) 
ஃ x = -1 and x = 1are vertical asymptotes,
Also
\(\underset { x-\rightarrow \infty }{ lim } \frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } =\underset { \frac { 1 }{ x } \rightarrow 0 }{ lim } \frac { 1 }{ 1-\frac { 1 }{ { x }^{ 2 } } } =1\)
[Divide numerator and denominator by x²]
ஃ y = 1 is a horizontal asymptote,