Freedom without limits is not freedom

Calculus Level 4

lim x ( x 4 + a x 3 + 3 x 2 + b x + 2 x 4 + 2 x 3 c x 2 + 3 x d ) \displaystyle \lim_{x \to \infty} \left ( \sqrt{x^4 + ax^3 + 3x^2 + bx + 2} - \sqrt{x^4 + 2x^3 - cx^2 + 3x - d} \ \right )

The limit above equals to 4 for constants a , b , c , d a,b,c,d .

What is the value of the expression below?

a + c + ( b 4 d 5 ) ( a 10 c ) a + c + \left ( b - \frac {4d}{5} \right ) \left ( a - \frac {10} c \right )


The answer is 7.

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2 solutions

Pi Han Goh
Mar 20, 2015

We multiply the expression in the limit by its "conjugate":

x 4 + a x 3 + 3 x 2 + b x + 2 + x 4 + 2 x 3 c x 2 + 3 x d x 4 + a x 3 + 3 x 2 + b x + 2 + x 4 + 2 x 3 c x 2 + 3 x d \frac {\sqrt{x^4 + ax^3 + 3x^2 + bx + 2} + \sqrt{x^4 + 2x^3 - cx^2 + 3x - d} }{\sqrt{x^4 + ax^3 + 3x^2 + bx + 2} + \sqrt{x^4 + 2x^3 - cx^2 + 3x - d} }

The limit becomes:

lim x ( x 4 + a x 3 + 3 x 2 + b x + 2 ) ( x 4 + 2 x 3 c x 2 + 3 x d ) x 4 + a x 3 + 3 x 2 + b x + 2 + x 4 + 2 x 3 c x 2 + 3 x d = lim x x 3 ( a 2 ) + x 2 ( c + 3 ) + x ( b 3 ) + ( d + 2 ) x 2 ( 1 + a x + 3 x 2 + b x 3 + 2 x 4 + 1 + 2 x c x 2 + 3 x 3 d x 4 ) \begin{aligned} & & \displaystyle \lim_{x \to \infty} \frac { (x^4 + ax^3 + 3x^2 + bx + 2) - (x^4 + 2x^3 - cx^2 + 3x - d)}{ \sqrt{x^4 + ax^3 + 3x^2 + bx + 2} + \sqrt{x^4 + 2x^3 - cx^2 + 3x - d} } \\ & = & \displaystyle \lim_{x \to \infty} \frac { x^3 (a-2) + x^2 (c+3) + x(b-3) + (d+2) } { x^2 \left ( \sqrt{1 + \frac {a}{x} + \frac {3}{x^2} + \frac {b}{x^3} + \frac {2}{x^4} } + \sqrt{1 + \frac { 2}{x} - \frac {c}{x^2} + \frac {3}{x^3} - \frac {d}{x^4} } \right ) } \\ \end{aligned}

Focusing on the denominator, for x x \to \infty , the denominator 2 x 2 \to 2x^2

If a 2 a \ne 2 , the the limit becomes O ( x 3 ) O ( x 2 ) \frac {O(x^3)}{O(x^2)} \to \infty , thus a = 2 a = 2

Now the expression of the limit is in the of O ( x 2 ) O ( x 2 ) \frac {O(x^2)}{O(x^2)} , with the ratio of coefficient of x 2 x^2 in the numerator and denominator as c + 3 2 \frac {c+3}{2} , which equals to the given value 4 4 , thus c = 5 c = 5 , and so

a 10 c = 2 10 5 = 0 ( b 4 d 5 ) ( a 10 c ) = 0 \large a - \frac {10}{c} = 2 - \frac {10} 5 = 0 \Rightarrow \left (b - \frac {4d}{5} \right ) \left (a - \frac {10}{c} \right) = 0

Hence, our answer is simply a + c = 2 + 5 = 7 a + c = 2 + 5 = \boxed{7}

Moderator note:

Wow that's long, I guess that's the only way to solve it. Good job!

We can also multiply above and below by 1/(x^2) ; take the 1/(x^2) in the numerator inside both the roots where it becomes 1/(x^4) ; then adjust the expression inside the roots to get a sort of "inverse polynomial" just like Pi Han Goh's 3rd step's denominator; and then apply L'Hospital's as it is 0/0 form.

However it is essentially the same ...Or is it ??

Santanu Banerjee - 6 years, 1 month ago

I have solved same vay. But c and d constats confused me at the first moment! :)

Ela Marinić-Kragić - 5 years, 9 months ago

I never checked for the required expression and went on the quest for finding b and d, which actually belong to R :P

Swapnil Das - 3 years, 1 month ago
Priyesh Pandey
Apr 11, 2015

using binomial expansion for rational index and neglecting higher degree terms one can easily evaluate this limit

Moderator note:

Can you elaborate more?

You have mint expansion series at x = x=\infty :

x 4 + a x 3 + 3 x 2 + b x + 2 x 4 + 2 x 3 c x 2 + 3 x + d = \sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x+d}=

= 1 2 ( a 2 ) x + ( a 2 8 + c 2 + 2 ) + a 3 12 a + 8 ( b c 4 ) 16 x + =\frac12(a-2)x+\left(-\frac{a^2}8+\frac c2+2\right)+\frac{a^3-12a+8(b-c-4)}{16x}+\dots ?

Ela Marinić-Kragić - 5 years, 9 months ago

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