Here the expression assumes the form ∞-∞ as x➝∞. So, we first reduce it to the rational form \(\cfrac { f(x) }{ g(x) } \) 
\(\underset { x\rightarrow \infty }{ lim } \sqrt { { x }^{ 2 }+x+1 } -\sqrt { { x }^{ 2 }+1 } =\underset { x\rightarrow \infty }{ lim } \cfrac { \left\{ \sqrt { { x }^{ 2 }+x+1 } -\sqrt { { x }^{ 2 }+1 } \right\} }{ \sqrt { { x }^{ 2 }+x+1 } +\sqrt { { x }^{ 2 }+1 } } \left\{ \sqrt { { x }^{ 2 }+x+1 } +\sqrt { { x }^{ 2 }+1 } \right\} \) 
= \(\underset { x\rightarrow \infty }{ lim } =\cfrac { { x }^{ 2 }+x+1-{ x }^{ 2 }-1 }{ \sqrt { { x }^{ 2 }+x+1 } +\sqrt { { x }^{ 2 }+1 } } \) 
= \(\underset { x\rightarrow \infty }{ lim } \cfrac { x }{ \sqrt { { x }^{ 2 }+x+1+\sqrt { { x }^{ 2 }+1 } } } =\underset { x\rightarrow \infty }{ lim } \cfrac { x }{ \sqrt { 1+\frac { 1 }{ x } +\cfrac { 1 }{ { x }^{ 2 } } +\sqrt { 1+\cfrac { 1 }{ { x }^{ 2 } } } } } \)
= \(\underset { x\rightarrow \infty }{ lim } \cfrac { 1 }{ \sqrt { 1+\frac { 1 }{ x } +\frac { 1 }{ { x }^{ 2 } } } +\sqrt { 1+\cfrac { 1 }{ { x }^{ 3 } } } } \)
= \(\cfrac { 1 }{ 1+1 } =\cfrac { 1 }{ 2 } \)

The given expression of the form ∞-∞ So we first we first write in the rational form \(\cfrac { f(x) }{ g(x) } \)
So that it reduces to either \(\cfrac { 0 }{ 0 } \) form \(\cfrac { \infty }{ \infty } \) form. 
\(\therefore \underset { x\rightarrow \infty }{ lim } \sqrt { x } \left\{ \sqrt { x+c } -\sqrt { x } \right\} =\underset { x\rightarrow \infty }{ lim } \cfrac { \sqrt { x } \left\{ \sqrt { x+c } -\sqrt { x } \right\} \left\{ \sqrt { x+c } +\sqrt { x } \right\} }{ \left\{ \sqrt { x+c } +\sqrt { x } \right\} } \) 
= \(\underset { x\rightarrow \infty }{ lim } \cfrac { \sqrt { x } \left( x+c-x \right) }{ \sqrt { x } +c+\sqrt { x } } \) 
= \(\underset { x\rightarrow \infty }{ lim } \cfrac { c\sqrt { x } }{ \sqrt { x+c } +\sqrt { x } } \) 
= \(\underset { x\rightarrow \infty }{ lim } \cfrac { c }{ \sqrt { 1+\frac { c }{ x } } +1 } \) 
= \(\underset { x\rightarrow \infty }{ lim } \cfrac { c }{ \sqrt { 1+\frac { c }{ x } } +1 } =\cfrac { c }{ 2 } \)

2x² -xy - 3y² - 6x + 19y - 20 = 0
Given equation of pair of lines is 2x² -xy - 3y² - 6x + 19y - 20 = 0
Consider 2x² -xy - 3y² = (2x - 3y) (x + y)
2x² - xy - 3y² - 6x + 19y- 20 = (2x - 3y + l) (x + y + m)

Equating the co-efficients of x and y terms we get,

 \(\Rightarrow\)  m = -5 
Substituting m = -5 in-6 = 2m + l we get,
-6 = 2(-5) + l \(\Rightarrow\) -6 = -10 + l \(\Rightarrow\) l = -6 +10 \(\Rightarrow\) l = 4
Hence the separate equations are 2x - 3y + 4 = 0 and  x + y - 5 = 0.

Parallel line to 3x + 4y = 7 is of the form 3x + 4y = 3x₁ + 4y₁. Let (x₁, y₁) be (1, 2)
⇒ 3x + 4y = 3(1) + 4(2)
3x + 4y = 11
Perpendicular line to 3x + 4y = 7 is of the form
4x - 3y = 4x₁ - 3y₁
Here (x₁, y₁) = (1, 2)
⇒ 4x - 3y = 4(1) - 3(2)
4x - 3y = -2
⇒ The parallel and perpendicular lines are respectively
3x + 4y = 1
4x - 3y = -2

Since the given equation is a homogeneous equation, divide the given equation 2x²+ 2xy + y² = 0 by x² and substituting \(\frac{y}{m}=m\)
We get m² + 2m + 2 = 0
The values of m (slopes) are not real (complex number), therefore no line will exist with the joint equation 2x² + 2xy +y² = 0
We sometimes say that this equation represents imaginary lines.
Note that in the entire plane, only (0, 0) satisfies this equation

We factorize this equation straight away as
5x² + 6xy + y² = 0
5x² + 5xy + xy + y²= 0
5x (x + y) + y (x + y) = 0
(5x +y) (x +y) = 0
So that the lines are 5x + y = 0, and x +y = 0
Alternate method :
since the given equation is a homogeneous equation, divide the given equation
5x² + 6xy + y² = 0 by x²
We get 5 + 6 \((\frac{x}{y} )+ (\frac{x}{y} )^2 = 0\)
Substitute \(\frac{y}{x}\) = m (slope of the lines for homogenous equation)
The above equation becomes m² + 6m + 5 = 0
Factorizing, we get
(m + 1) (m + 5) = 0
m = −1, m = −5
\((\frac{x}{y} )=1, (\frac{x}{y} )^2 = 5\)
That is, the lines are x + y = 0, 5x + y = 0

Let p be the length of the perpendicular drawn from the origin to the required line.
The perpendicular makes 45° with the x-axis.
The equation of the required line is of the form,
x cos \(\alpha\) + y sin \(\alpha\) = p
⇒ x cos 45° +y sin 45° = p
x + y = √2p
This equation cuts the coordinate axes at A(√2p, 0) and B(0, √2p)
Area of the ΔOAB is \(\frac{1}{2}\times\sqrt{2}p\times\sqrt{2}p=36\)
p = 6
Therefore the equation of the required line is x + y = 6\(\sqrt{2}\)

There are two straight lines making 30° with the y-axis.
From the figure, it is clear that the two lines make the angles 60° and 120° with the x-axis
Let m₁ be tan 60° = √3 and 
m₂ be tan 120° = tan( 180° - 60°)

= -tan60°= -√3
m₁ = √3, m₂ = -√3 and b = 7
Equations of lines are y = m₁x + b and y = m₂x + b
y = √3x + 7 and y = -√3x + 7

Let A, B and C be \((0,-\frac{3}{2}),(1,-1)\) and \((2,-\frac{1}{2})\)  respectively.
The slope of AB is \(\frac{-1+\frac{3}{2}}{1-0}=\frac{1}{2}\)
The slope of BC is \(\frac{\frac{-1}{2}+1}{2-1}=\frac{1}{2}\)
Thus, the slope of AB is equal to slope of BC.
Hence, A, Band C are lying on the same line.

Here, p = 12 and \(\alpha\) = 150°, So the equation of the required line is of the form
x cos \(\alpha\) + y sin \(\alpha\) = p
That is, x cos150° +y sin150° = 12
⇒ √3x - y + 24 = 0