Evaluating the limit $\lim_{x \to \infty}\left(\sqrt{x^4 -x^3+1}-\sqrt{x^4+15x^2-5}\,\right)$

Question

I wanna know how to do this limit

$\lim_{x \to \infty}\left(\sqrt{x^4 -x^3+1}-\sqrt{x^4+15x^2-5}\,\right)$

Answer 1

First of all, rationalise the expression by multiplying numerator and denominator by$\displaystyle \sqrt{x^{4} -x^{3} +1} \ +\ \sqrt{x^{4} +15x^{2} -5}$ \begin{equation*} \end{equation*} We get \begin{gather*} \lim x\rightarrow \infty \ \frac{-x^{3} -15x^{2} +6}{\sqrt{x^{4} -x^{3} +1} \ +\ \sqrt{x^{4} +15x^{2} -5}}\\ \end{gather*} Divide numerator and denominator by $\displaystyle x^{2}$ \begin{equation*} \lim x\rightarrow \infty \ \frac{-x-15+\frac{6}{x^{2}}}{\sqrt{1-\frac{1}{x} +\frac{1}{x^{4}}} +\sqrt{1+\frac{15}{x^{2}} -\frac{5}{x^{4}}}} \end{equation*}

Clearly the given expression tends to$\displaystyle \ -\infty $ as $\displaystyle x\rightarrow \infty $ \begin{equation*} \end{equation*}

Answer 2

You should multiply by the conjugate of the expression, then proceed.

\begin{align} \frac{\left(\sqrt{x^4 -x^3+1}-\sqrt{x^4+15x^2-5}\,\right)\left(\sqrt{x^4 -x^3+1}+\sqrt{x^4+15x^2-5}\,\right)}{\left(\sqrt{x^4 -x^3+1}+\sqrt{x^4+15x^2-5}\,\right)}\\ \end{align} \begin{align} \frac{(x^4 -x^3+1)-(x^4+15x^2-5)}{\left(\sqrt{x^4 -x^3+1}+\sqrt{x^4+15x^2-5}\,\right)} \end{align}

Operate and calculate the limit.

Answer 3

Another manipulation:

Let $a,b >0$, real.

$a-b=(√a-√b)(√a+√b)$, then

$√a-√b = \dfrac{a-b}{√a+√b}.$

$a: = (x^4-x^3+1)^{1/2},$ $b=(x^4+15x^2-5)^{1/2}.$

$\small{f(x):=}$

$\small {=\dfrac{-x^3-15x^2+6}{(x^4-x^3+1)^{1/2}+(x^4+15x^2-5)^{1/2}}}$

Note : For large x (positive) the numerator is negative, the denominator positive: Increasing the denominator makes the fraction bigger.

Hence

$f(x) \lt $

$\dfrac{-x^3+6}{(x^4+1)^{1/2}+(x^4+15x^2)^{1/2}} \lt $

$\dfrac{-x^3+6}{√2x^2+√2x^2}=$

$\dfrac{-x^3+6}{(2√2)x^2}.$

Take the limit.