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

I wanna know how to do this limit

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

• Try multiplying and dividing by the conjugate root to eliminate the roots.
– D.B.
Commented Nov 22, 2018 at 4:37
• thats what i did, but i got stuck Commented Nov 22, 2018 at 4:41
• math.stackexchange.com/questions/2959619/… Commented Nov 22, 2018 at 4:41
• Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
– Ovi
Commented Nov 22, 2018 at 4:49

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 $$\displaystyle x\rightarrow \infty$$ $$\begin{equation*} \end{equation*}$$

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.

• negative! that's cool! Commented Nov 22, 2018 at 4:52
• What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse. Commented Nov 22, 2018 at 4:54
• and then, should i evaluate de sign of de function to see the lateral limits? Commented Nov 22, 2018 at 4:54
• No need if you just want to give a result of the limit. If you want to study the function then I suppose you should Commented Nov 22, 2018 at 4:56
• Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor? Commented Nov 22, 2018 at 4:56

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.