Find: $\lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)

Question

Find: $\displaystyle \lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)

After developing the expression, by multiplying the fraction by the conjugates and rearranging, I found:

$$f(x)=x\times \frac{1+\sqrt{1+1/x^6}}{1+\sqrt{1+1/x^4}}$$

I can solve it for the limit when $x\to \infty$, with answer $\infty$ which is correct (by the book and Wolfram Alpha). But when the limit is $x\to -\infty$ the answer is zero, but from this expression, I get $-\infty$ as the answer.

I'm certainly missing something.

Hints and answers appreciated. Sorry if this is a duplicate.

Answer 1

Set $x=-y\to+\infty$

$$\frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}=\frac{y^2-\sqrt{y^4+1}}{-y^3-\sqrt{y^6+1}}=-\frac{y^2-\sqrt{y^4+1}}{y^3+\sqrt{y^6+1}}\frac{y^2+\sqrt{y^4+1}}{{y^2+\sqrt{y^4+1}}}=\frac{1}{(y^3+\sqrt{y^6+1})({y^2+\sqrt{y^4+1})}}\to0$$

Answer 2

If $x\to-\infty,$ let $-1/x=h,h\to0^+$

$$\sqrt{x^{2n}+1}=\dfrac{\sqrt{1+h^{2n}}}{\sqrt{h^{2n}}}=\dfrac{\sqrt{1+h^{2n}}}{h^n}$$ as $h^n>0$ as $h>0$

$$\lim_{x\to-\infty}\dfrac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}=-\lim_{h\to0^+}h\cdot\dfrac{1-\sqrt{1+h^4}}{1+\sqrt{1+h^6}}=0$$

Answer 3

$$\frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}=\frac{x^4-x^4-1}{x^6-x^6-1}\cdot \frac{x^3+\sqrt{x^6+1}}{x^2+\sqrt{x^4+1}}\to \frac{x^3+|x^3|}{x^2+|x^2|}=\frac{x^3-x^3}{x^2+x^2}= 0$$

Answer 4

Note that if $x\lt 0$, we have $$\sqrt{x^6\left(1+\frac{1}{x^6}\right)}=|x^3|\sqrt{1+\frac{1}{x^6}}=\color{red}{-x^3}\sqrt{1+\frac{1}{x^6}}$$

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$$\begin{align}&\lim_{x\to -\infty}\frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}\\\\&=\lim_{x\to -\infty}\frac{(x^2-\sqrt{x^4+1})(x^2+\sqrt{x^4+1})(x^3+\sqrt{x^6+1})}{(x^3-\sqrt{x^6+1})(x^2+\sqrt{x^4+1})(x^3+\sqrt{x^6+1})}\\\\&=\lim_{x\to -\infty}\frac{x^3+\sqrt{x^6+1}}{x^2+\sqrt{x^4+1}}\\\\&=\lim_{x\to -\infty}\frac{x^3+\sqrt{x^6(1+\frac {1}{x^6})}}{x^2+\sqrt{x^4(1+\frac{1}{x^4})}}\\\\&=\lim_{x\to -\infty}\frac{x^3+|x^3|\sqrt{1+\frac{1}{x^6}}}{x^2+|x^2|\sqrt{1+\frac{1}{x^4}}}\\\\&=\lim_{x\to -\infty}\frac{x^3+(\color{red}{-x^3})\sqrt{1+\frac{1}{x^6}}}{x^2+x^2\sqrt{1+\frac{1}{x^4}}}\\\\&=\lim_{x\to -\infty}x\cdot\frac{\left(1-\sqrt{1+\frac{1}{x^6}}\right)\left(1+\sqrt{1+\frac{1}{x^6}}\right)}{\left(1+\sqrt{1+\frac{1}{x^4}}\right)\left(1+\sqrt{1+\frac{1}{x^6}}\right)}\\\\&=\lim_{x\to -\infty}\frac{-1}{x^5\left(1+\sqrt{1+\frac{1}{x^4}}\right)\left(1+\sqrt{1+\frac{1}{x^6}}\right)}=0\end{align}$$

Answer 5

Multiplying by one of its conjugates we get: $\displaystyle \lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}=\displaystyle \lim_{x\to -\infty} \frac{-1}{\big(x^2+\sqrt{x^4+1}\big)\big(x^3-\sqrt{x^6+1}\big)}=0.$

Notice that in the second term we have used the fact that $\sqrt{x^6}=|x^3|=-x^3$.

Answer 6

Let $y=-x:$

$F(y) = \dfrac{y^2 -\sqrt{y^4 +1}}{-y^3 - \sqrt{y^6+1}}=$

$(-1)\dfrac{y^2 - \sqrt{y^4+1}}{y^3 +\sqrt{y^6+1}}.$

Conjugates and rearrangement , and then considering $\lim_{y \rightarrow \infty}$ should get you there .

Or:

$0\le |F(y)| \le \dfrac{2\sqrt{y^4+1}}{2y^3} \lt \dfrac{4y^2}{2y^3}$.

And the limit is?.

Used:

$ |y^2 - \sqrt{y^4+1}| \lt 2\sqrt{y^4+1} \lt 4y^2.$

Answer 7

You certainly forgot an absolute value in your computations.

Here is a computation for $x<0$: \begin{align} f(x)&= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}=-\frac 1{\Bigl(x^3-\sqrt{x^6+1}\Bigr)\Bigl(x^2+\sqrt{x^4+1}\Bigr)}\\ &= -\frac 1{\biggl(x^3-|x^3|\sqrt{1+\dfrac1{x^6}}\mkern2mu\biggr)\Bigl(x^2+\sqrt{x^4+1}\Bigr)} \sim_{-\infty} \frac{-1}{2x^3\cdot2x^2}=-\frac1{4x^5}\to 0^+. \end{align}