Question

If $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^6} + a{x^5} + b{x^3} - cx + d} - \sqrt {{x^6} - 2{x^5} + {x^3} + 1} } \right) = 2$ then
A) b=$- 2$
B) a=$- 3$
C) a=$2$
D) b=$- 3$

Answer 1

Rationalize the given function and put $\dfrac{1}{x} = 0$ in the obtained result because $x \to - \infty$ .Then sove to find the value of a and b.

Complete step-by-step answer:
Given function is $\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^6} + a{x^5} + b{x^3} - cx + d} - \sqrt {{x^6} - 2{x^5} + {x^3} + x + 1} } \right) = 2$
It is in $\infty - \infty$ indeterminate form. So we can solve it by rationalizing.
On rationalizing the given function we get,
$\Rightarrow \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {\sqrt {{x^6} + a{x^5} + b{x^3} - cx + d} - \sqrt {{x^6} - 2{x^5} + {x^3} + 1} } \right) \times \left( {\sqrt {{x^6} + a{x^5} + b{x^3} - cx + d} + \sqrt {{x^6} - 2{x^5} + {x^3} + x + 1} } \right)}}{{\left( {\sqrt {{x^6} + a{x^5} + b{x^3} - cx + d} + \sqrt {{x^6} - 2{x^5} + {x^3} + x + 1} } \right)}} = 2$ We know that ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ , On applying this in the above equation we get,
$\Rightarrow$ $\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {{{\left( {\sqrt {{x^6} + a{x^5} + b{x^3} - cx + d} } \right)}^2} - {{\left( {\sqrt {{x^6} - 2{x^5} + {x^3} + x + 1} } \right)}^2}} \right)}}{{\left( {\sqrt {{x^6} + a{x^5} + b{x^3} - cx + d} + \sqrt {{x^6} - 2{x^5} + {x^3} + x + 1} } \right)}} = 2$
On solving we get,
$\Rightarrow \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^6} + a{x^5} + b{x^3} - cx + d - \left( {{x^6} - 2{x^5} + {x^3} + x + 1} \right)}}{{\left( {\sqrt {{x^6} + a{x^5} + b{x^3} - cx + d} + \sqrt {{x^6} - 2{x^5} + {x^3} + x + 1} } \right)}} = 2$
On multiplying the negative sign inside the bracket,
$\Rightarrow \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^6} + a{x^5} + b{x^3} - cx + d - {x^6} + 2{x^5} - {x^3} - x - 1}}{{\left( {\sqrt {{x^6} + a{x^5} + b{x^3} - cx + d} + \sqrt {{x^6} - 2{x^5} + {x^3} + x + 1} } \right)}} = 2$
On taking the coefficients of the same terms common we get,
$\Rightarrow \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {a + 2} \right){x^5} + \left( {b - 1} \right){x^3} - \left( {c + 1} \right)x + d - 1}}{{\left( {\sqrt {{x^6} + a{x^5} + b{x^3} - cx + d} + \sqrt {{x^6} - 2{x^5} + {x^3} + x + 1} } \right)}} = 2$
On taking ${x^6}$ common in the denominator we get,
$\Rightarrow \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {a + 2} \right){x^5} + \left( {b - 1} \right){x^3} - \left( {c + 1} \right)x + d - 1}}{{{x^3}\left( {\sqrt {1 + \dfrac{a}{x} + \dfrac{b}{{{x^3}}} - \dfrac{c}{{{x^5}}} + \dfrac{d}{{{x^6}}}} + \sqrt {1 - \dfrac{2}{x} + \dfrac{1}{{{x^3}}} + \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} } \right)}} = 2$
On taking ${x^3}$ common from numerator we get,
$\Rightarrow \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^3}\left[ {\left( {a + 2} \right){x^2} + \left( {b - 1} \right) - \dfrac{{\left( {c + 1} \right)}}{{{x^2}}} + \dfrac{{d - 1}}{{{x^3}}}} \right]}}{{{x^3}\left( {\sqrt {1 + \dfrac{a}{x} + \dfrac{b}{{{x^3}}} - \dfrac{c}{{{x^5}}} + \dfrac{d}{{{x^6}}}} + \sqrt {1 - \dfrac{2}{x} + \dfrac{1}{{{x^3}}} + \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} } \right)}} = 2$
On simplifying we get,
$\Rightarrow \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left[ {\left( {a + 2} \right){x^2} + \left( {b - 1} \right) - \dfrac{{\left( {c + 1} \right)}}{{{x^2}}} + \dfrac{{d - 1}}{{{x^3}}}} \right]}}{{\left( {\sqrt {1 + \dfrac{a}{x} + \dfrac{b}{{{x^3}}} - \dfrac{c}{{{x^5}}} + \dfrac{d}{{{x^6}}}} + \sqrt {1 - \dfrac{2}{x} + \dfrac{1}{{{x^3}}} + \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} } \right)}} = 2$
Now since it is given that $x \to - \infty \Rightarrow \dfrac{1}{x} = 0$
So all the values multiplied with $\dfrac{1}{x}$ in the denominator and numerator will be zero and for the limit to be finite as x tends to infinite the quantity $a - 2 = 0$ $\Rightarrow a = 2$ .So the limit will only exist for $b - 1$
Then on putting the values we get,
$\Rightarrow \dfrac{{\left[ {\left( {b - 1} \right)} \right]}}{{\left( {\sqrt 1 + \sqrt 1 } \right)}} = 2$
On simplifying we get,
$\Rightarrow \dfrac{{b - 1}}{2} = 2 \\\ \Rightarrow b - 1 = 4 \\\ \Rightarrow b = 5 \\\$

So, only option C is correct.

Note: Here we rationalize the function to make it easier to solve the limit as the function gives indeterminate form on putting the limit. Indeterminate forms are such forms which cannot be determined so we try to find another method to change the indeterminate form. Here we have used rationalization. In some questions, we use L’ Hospital rule which states that the limit of derivative of the given function is equal to the limit of indeterminate form.