Question

Evaluate $\mathop {\lim }\limits_{x \to \infty } \left[ {\dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{x + 1}}{{{x^2} + 1}}} \right]$

Answer 1

Apply L'Hospital's rule to evaluate the equation as we have an indeterminate form $\mathop {\lim }\limits_{x \to \infty }$ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit. The limit when we divide one function by another is the same after we take the derivative of each function.

Complete step by step solution:
Let us write the given input to evaluate
$\mathop {\lim }\limits_{x \to \infty } \left[ {\dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{x + 1}}{{{x^2} + 1}}} \right]$
The limit of a $\dfrac{{sum}}{{difference}}$is the $\dfrac{{sum}}{{difference}}$of limits:
$\mathop {\lim }\limits_{x \to \infty } \left[ {\dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{x + 1}}{{{x^2} + 1}}} \right]$= $\left[ {\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \mathop {\lim }\limits_{x \to \infty } \dfrac{{x + 1}}{{{x^2} + 1}}} \right]$
Multiply and divide by ${x^2}$:
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \mathop {\lim }\limits_{x \to \infty } \dfrac{{x + 1}}{{{x^2} + 1}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \mathop {\lim }\limits_{x \to \infty } \dfrac{{{x^2}\dfrac{{x + 1}}{{{x^2}}}}}{{{x^2}\dfrac{{{x^2} + 1}}{{{x^2}}}}}$
Divide:
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \mathop {\lim }\limits_{x \to \infty } \dfrac{{x + 1}}{{{x^2} + 1}}$ = $\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{1}{x} + \dfrac{1}{{{x^2}}}}}{{1 + \dfrac{1}{{{x^2}}}}}$
The limits of the quotient are the quotient of limits:
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{1}{x} + \dfrac{1}{{{x^2}}}}}{{1 + \dfrac{1}{{{x^2}}}}}$ = $\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{x} + \dfrac{1}{{{x^2}}}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{1}{{{x^2}}}} \right)}}$
The limit of a $\dfrac{{sum}}{{difference}}$is the $\dfrac{{sum}}{{difference}}$of limits is
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{x} + \dfrac{1}{{{x^2}}}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{1}{{{x^2}}}} \right)}} + \dfrac{{\left( {\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{{{x^2}}} + \mathop {\lim }\limits_{x \to \infty } \dfrac{1}{x}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{1}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}}$
The limit of a quotient is the quotient of limits:
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{x} + \mathop {\lim }\limits_{x \to \infty } \dfrac{1}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{1}{{{x^2}}}} \right)}} + \dfrac{{\left( {\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{{{x^2}}} + \dfrac{{\mathop {\lim }\limits_{x \to \infty } 1}}{{\mathop {\lim x}\limits_{x \to \infty } }}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{1}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}}$
The limit of a constant is equal to the constant:
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{\dfrac{{\mathop {\lim }\limits_{x \to \infty } 1}}{{\mathop {\lim }\limits_{x \to \infty } x}} + \mathop {\lim }\limits_{x \to \infty } \dfrac{1}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{1}{{{x^2}}}} \right)}} + \dfrac{{\left( {\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{{{x^2}}} + \dfrac{{\mathop {\lim }\limits_{x \to \infty } 1}}{{\mathop {\lim x}\limits_{x \to \infty } }}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{1}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}}$
Constant divided by maximum number which equals 0:
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{1\dfrac{1}{{\mathop {\lim }\limits_{x \to \infty } x}} + \mathop {\lim }\limits_{x \to \infty } \dfrac{1}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{1}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{{{x^2}}} + 0}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{1}{{{x^2}}}} \right)}}$
The limit of a quotient is the quotient of limits:
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{1}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{\dfrac{{\mathop {\lim }\limits_{x \to \infty } 1}}{{\mathop {\lim {x^2}}\limits_{x \to \infty } }}}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 + \dfrac{1}{{{x^2}}}} \right)}}$
The limit of a constant is equal to the constant:
$\mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{\mathop {\lim }\limits_{x \to \infty } 1}}{{\mathop {\lim }\limits_{x \to \infty } {x^2}\left( {1 + \dfrac{1}{{{x^2}}}} \right)}} + \dfrac{1}{{\mathop {\lim }\limits_{x \to \infty } {x^2}\left( {1 + \dfrac{1}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}}$
Divide by highest denominator power:
= $\dfrac{{2 - \dfrac{1}{x} - \dfrac{6}{{{x^2}}}}}{{3 + \dfrac{2}{x} + \dfrac{1}{{{x^2}}}}}$
= $\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{2 - \dfrac{1}{x} - \dfrac{6}{{{x^2}}}}}{{3 + \dfrac{2}{x} + \dfrac{1}{{{x^2}}}}} + \dfrac{{x + 1}}{{{x^2} + 1}}} \right)$

= $\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{2 - \dfrac{1}{x} - \dfrac{6}{{{x^2}}}}}{{3 + \dfrac{2}{x} + \dfrac{1}{{{x^2}}}}}} \right) + \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{x + 1}}{{{x^2} + 1}}} \right)$
$\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{2 - \dfrac{1}{x} - \dfrac{6}{{{x^2}}}}}{{3 + \dfrac{2}{x} + \dfrac{1}{{{x^2}}}}}} \right)$ = $\dfrac{2}{3}$
And
$\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{x + 1}}{{{x^2} + 1}}} \right)$= 0
Hence, after simplifying we get the value as
= $\dfrac{2}{3} + 0$
= $\dfrac{2}{3}$
Therefore, after evaluating all the terms we get
$\mathop {\lim }\limits_{x \to \infty } \left[ {\dfrac{{2{x^2} - x - 6}}{{3{x^2} + 2x + 1}} + \dfrac{{x + 1}}{{{x^2} + 1}}} \right]$ = $\dfrac{2}{3}$

Additional information:
Let us know the statement of L'Hospital's rule defines:
It says that the limit when we divide one function by another is the same after we take the derivative of each function
It is denoted as:
$\mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}}$

Formula used:
$\mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}}$
As where the limit of x is till infinity.

Note:
For a limit approaching the given value, the original functions must be differentiable on either side of value, but not necessarily at the value given. The limit of a quotient is equal to the quotient of the limits. The limit of a constant function is equal to the constant. The limit of a linear function is equal to the number x is approaching.