\(\underset { x\rightarrow \infty }{ lim } { e }^{ -x }\sqrt { x } =\underset { x\rightarrow \infty }{ lim } \frac { { \sqrt { x } } }{ { e }^{ x } } =\frac { \infty }{ \infty } \)
Which is in indeterminate form. Applying L' Hopital rule we get,
\(\underset { x\rightarrow \infty }{ lim } \frac { \frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } -1 } }{ { e }^{ x } } =\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } \frac { { x }^{ \frac { 1 }{ 2 } -1 } }{ { e }^{ x } } =\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } \frac { { e }^{ -x } }{ \sqrt { x } } \)
\(\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } e^{ -x }\sqrt { \frac { 1 }{ x } } =\frac { 1 }{ 2 } { e }^{ -\infty }(0)=0\) [When x ➝ ∞, \(\frac1x\) ➝ 0 e^-∞ = 0]

This is an indeterminate form \(\frac{\infty}{\infty}\) and hence we use the l’Hôpital’s Rule to evaluate.
\(\underset{x\rightarrow 1^{-}}{lim}(\frac{log(1-x)}{cot(\pi x)})=\underset{x\rightarrow1^{-}}{lim}(\frac{-\frac{1}{1-x}}{-\pi cosec^{2}(\pi x)})\) \((\frac{\infty}{\infty})\) 
On Simplication,
\(=\underset{x\rightarrow-1}{lim}(\frac{sin^{2}(\pi x)}{\pi (1-x)})\) \((\frac{0}{0})\)
again applying the l’Hôpital Rule 
\(= \underset{x\rightarrow 1^{-}}{lim}(\frac{2\pi sin(\pi x). cos(\pi x)}{-\pi})\)
=\(\underset{x\rightarrow -1}{lim}(-2 sin(\pi x).cos (\pi x))\)
= 0.

Let (x) = sin x
f^I(x) = cos x
f^II(x) = - sin x
f^III(x) = -cos x
⇒ \(f\left( \frac { \pi }{ 4 } \right) =sin\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)
\({ f }^{ 1 }\left( \frac { \pi }{ 4 } \right) =cos\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)
\({ f }^{ II }\left( \frac { \pi }{ 4 } \right) =sin\frac { \pi }{ 4 } =-\frac { 1 }{ \sqrt { 2 } } \)
\({ f }^{ II }\left( \frac { \pi }{ 4 } \right) =-cos\frac { \pi }{ 4 } =-\frac { 1 }{ \sqrt { 2 } } \)
Taylors' series for f (x) at x = \(\frac { \pi }{ 4 } \) is
\(f(x)=f\left( \frac { \pi }{ 4 } \right) +\frac { { f }^{ I }\left( \frac { \pi }{ 4 } \right) }{ 1! } \left( x-\frac { \pi }{ 4 } \right) +\frac { { f }^{ II }\left( \frac { \pi }{ 4 } \right) }{ 2! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 2 }+\frac { { f }^{ III }\left( \frac { \pi }{ 4 } \right) }{ 3! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 3 }\)+...
\(sinx=\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \left( x-\frac { \pi }{ 4 } \right) -\frac { 1 }{ \sqrt { 2 } } \frac { { \left( x-\frac { \pi }{ 4 } \right) }^{ 2 } }{ 2! } -\frac { 1 }{ \sqrt { 2 } } \frac { { \left( x-\frac { \pi }{ 4 } \right) }^{ 3 } }{ 3! } +...\)
\(=\frac { 1 }{ \sqrt { 2 } } \left[ 1+\left( x-\frac { \pi }{ 4 } \right) -\frac { 1 }{ 2! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 2 }-\frac { 1 }{ 3! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 3 }+..... \right] \)
\(sinx=\frac { \sqrt { 2 } }{ 2 } \left[ 1+\frac { 1 }{ 1! } \left( x-\frac { \pi }{ 4 } \right) -\frac { 1 }{ 2! } { \left( x-\frac { \pi }{ 4 } \right) }-\frac { 1 }{ 3! } { \left( x-\frac { \pi }{ 4 } \right) }+..... \right] \)

Let f(x) = log x
f^l(x) = \(\frac{1}{x}\) = x^-1
f^lI(x) = -1x^-2
f^III(x) = 2x^-3
f^IV(x) = -6 x^-4
⇒ f(1) = log 1 = 0
f^l(1) = \(\frac11\) = 1
f^ll(1) = -1
f^lll(1) = 2
f^IV(1) = -6
Taylor series for f(x) at x = 1 is
\(f(x)=f(1)+\frac { { f }^{ 1 }(1) }{ 1! } (x-1)+\frac { { f }^{ II }(1) }{ 2! } ({ x-1) }^{ 2 }+\frac { { f }^{ III }(1) }{ 3! } { (x-1) }^{ 3 }+\).....
log x = \(\frac { 1 }{ 1! } (x-1)+\frac { -1 }{ 2! } { (x-1) }^{ 2 }+\frac { 2 }{ 3! } ({ x-1) }^{ 3 }-\frac { 6 }{ 4! } ({ x-1) }^{ 4 }+..\)
log x = \((x-1)-\frac { 1 }{ 2 } { (x-1) }^{ 2 }+\frac { 1 }{ 3 } { (x-1) }^{ 3 }-\frac { 1 }{ 4 } ({ x-1) }^{ 4 }\)+ ....

Given f(x) = Ax² + Bx + C, x ∈ [a, b]
a) f (x) is continuous in [a, b]
b) f(x) is differentiate in (a, b)
c) f(b) = Ab²+ Bb + c,
f(a) = Aa²+ Ba + c
Using mean value theorem, there exists c ∈ [a, b] such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
2Ac + B = \(\frac { (A{ b }^{ 2 }+Bb+C)-(A{ a }^{ 2 }+Ba+C) }{ b-a } \)
\(=\frac{A b^{2}+B b+\not C-A a^{2}-B a-\not C}{b-a}\)
= \(\frac { a(b+a)(b-a)+B(b-a) }{ b-a } \)
\(=\frac{(\not b-a)[\mathrm{A}( b+a)+\mathrm{B}]}{\not b-a}\)
\(2 \mathrm{~A} c+\not \mathbf{B}=\mathrm{A}(a+b)+\not \mathrm{B}\)
⇒ 2Ac = A(a + b)
= \(\frac{a+b}{2}\) ∈ [a, b]

f(x) = x² -12x + 10; [1, 2]
Given f(x) = x² -12x + 10 ; [1, 2]
f'(x) = 2x - 12
f'(x) = 0
\(\Rightarrow\) 2x-12 = 0
\(\Rightarrow\) 2x = 12
\(\Rightarrow\) x = 6
\(\therefore\) The critical number is 6
Evaluating f (x) at the end points x = 1,
x = 2 and at the critical number x = 6 we get
f(1) = 1²-12(1)+10 = -1
f(2) = 2²-12(2)+10 = -10
Absolute maximum f(1) = -1
Absolute minimum f(2) = -10

The given function is a periodic function with period 2π and hence there will be stationary points and points of inflections in each period interval. We have,

\(\frac { dy }{ dx } =cosx\) and \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =sinx\)
Now,\( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-sinx=0\Rightarrow x=n\pi \)
We now consider an interval, (-π, π ) by splitting into two sub intervals \(\left( -\pi ,0 \right) \) and \(\left( 0,\pi \right) \)
In the interval \(\left( -\pi ,0 \right) ,\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } >0\) and hence the function is concave upward
In the interval \(\left( -\pi ,0 \right) ,\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } <0\) and hence the function is concave downward. Therefore (0,3) is a point of inflection. The general intervals need to be considered to discuss the concavity of the curve are \(\left( n\pi ,\left( n+1 \right) \pi \right) \), where n is any integer which can be discussed as before to conclude that \(\left( n\pi ,3 \right) \) are also points of inflection.

 We have , f(x) = \(x^\frac{2}{3}\) then \(f'(x)=\frac{2}{3}x^{-\frac{1}{3}}\)\(​​​​=\frac{2}{3x^{\frac{1}{3}}}, f'(x)\ne 0 \forall x \in R\) and f'(x) does not exist at x = 0.
Therefore, there are no stationary points but there is a critical point at x = 0.

Interval	\((-\infty,0)\)	\((0,\infty)\)
Sign of f'(x)	-	+
Monotonicity	strictly decreasing	strictly increasing
	\(\searrow \)	↗️

Because f'(x) changes its sign from negative to positive when passing through x = 0 for the function it has a local minimum at x = 0. The local minimum value is f(0) = 0. Note that here the local minimum occurs at a critical point which is not a stationary point.

Since the polynomial in the numerator is a higher degree (2^nd) than the denominator (1^st), we know we have a slant asymptote. To find it, we must divide the numerator by the denominator. We can use long division to do that:
\(\begin{array}{r} x-11 \\ x + 5 \sqrt { x ^ { 2 } - 6 x + 7 } \\ \frac{x^{2}+5 x}{-11 x+7} \\ \frac{-11 x-55}{62} \end{array}\)
Notice that we don't need to finish the long division problem to find the remainder. We only need the terms that will make up the equation of the line. The slant asymptote is y = x −11.

As you can see in this graph of the function, the curve approaches the slant asymptote y = x −11 but never crosses it:

We have,
\(\underset { x\rightarrow { 0 }^{ - } }{ lim } =-\infty \ and\ \underset { x\rightarrow { 0 }^{ x } }{ lim } =\frac { 1 }{ x } =\infty \). Hence, the required vertical asymptote is x = 0 or the y -axis.
As the curve is symmetric with respect to both the axes, y = 0 or the x -axis is also an asymptote.
Hence this (rectangular hyperbola) curve has both the vertical and horizontal asymptotes.