The limit of a function with sum of two roots.

Question

I need to find the limit of the following function: $$\lim_{x\to-\infty} \left(\sqrt{x^2+2x}+\sqrt[3]{x^3+x^2}\right).$$ I derived it to the form of: $$\lim_{x\to-\infty}\frac{(x^2+2x)^3-(x^3+x^2)^2}{\left(\sqrt{x^2+2x}-\sqrt[3]{x^3+x^2}\right)\left((x^2+2x)^2+(x^2+2x)\sqrt[3]{(x^3+x^2)^2}+\sqrt[3]{(x^3+x^2)^4}\right)},$$ hoping that I'd be able to simplify something but still have $\frac{0}{0}$ and don't see how to do that.

Answer 1

$$\lim_{x\to-\infty}\sqrt{x^2+2x}\color{red}{+}\sqrt[3]{x^3+x^2}\\=\lim_{x\to-\infty}|x|\left(1+{2\over x}\right)^{1\over 2}\color{red}{+}x\left(1+{1\over x}\right)^{1\over3}\\=\lim_{x\to-\infty}|x|\left(1+{1\over 2}\cdot{2\over x}\right)\color{red}{+}x\left(1+{1\over 3}\cdot{1\over x}\right)=-{2\over 3}$$

NB:

$\lim_{n\to\infty}n(1+{1\over n})^p$ can be written as $n(1+{p\over n})$ as the other terms will be like ${1\over n},{1\over n^2}$ etc which can be neglected.($p$ is finite)

Answer 2

Let me change the sign on $x$, I find it easier to think about. the expression is then

$$\sqrt{x^2-2x}+\sqrt[3]{x^2-x^3}$$

Now investigate the two limits $$\sqrt{x^2-2x}-x\rightarrow -1$$

and

$$x+\sqrt[3]{x^2-x^3}$$ and this latter is equal

to

$$\frac{x^2}{x^2-x\sqrt[3]{x^2-x^3}+(\sqrt[3]{x^2-x^3})^2}$$

for which I get the limit of $\frac{1}{3}$ (assuming I haven't made a mistake, (a big assumption)).

Ok so I get the limit $-\frac{2}{3}$ in agreement with the book.

Answer 3

For $x<0$,

$$f(x)=-x(1+\frac{2}{x})^{\frac{1}{2}}+x(1+\frac{1}{x})^\frac{1}{3}$$

$$=-x-1+x+\frac{1}{3}+\frac{1}{x}\epsilon(x).$$

the limit is $$-\frac{2}{3}.$$

Answer 4

When $x$ is negative, then $\sqrt{x^2} = -x$, which is positive. So we have

\begin{align} & \lim_{x\to-\infty} \left(\sqrt{x^2+2x}+\sqrt[3]{x^3+x^2}\right) \\[10pt] = {} & \lim_{x\to-\infty} \left( -x\sqrt{1 + \frac 2 x} + x \sqrt[3]{1 + \frac 1 x} \, \right) \\[10pt] = {} & \lim_{x\to-\infty} x\left( -\left( 1 + \frac 2 x \right)^{1/2} + \left( 1 + \frac 1 x \right)^{1/3} \right) \\[10pt] = {} & \lim_{x\to-\infty} x\left( -\left( 1 + \frac 2 x \right)^{1/2} + \left( 1 + \frac 1 x \right)^{1/3} \right) \\[10pt] = {} & \lim_{u\,\uparrow\, 0} \frac 1 u \left( - (1+2u)^{1/2} + (1+u)^{1/3} \right) \text{ where } u = \frac 1 x. \end{align} L'Hopital's rule can deal with that. (And remember the chain rule when finding $\dfrac d {du} (1+2u)^{1/2}.$)