3

$$\mathop {\lim }\limits_{x \to \infty } (\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} ) $$

My try:

$${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2}) $$

$$\mathop {\lim }\limits_{x \to \infty } \frac{{(\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} )(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}}{{(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}} = $$

$$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} + 5{x^2} - {x^2} - 2x}}{{(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}} $$

And what's next...?

This task in first and second remarkable limits. I think i can replace variable, but how i will calculate it...

Martin Sleziak
  • 53,687
Evgeny Semyonov
  • 825
  • 4
  • 12

4 Answers4

2

If you can use the Taylor expansion $(1+t)^\alpha=1+\alpha t + \hbox{higher order terms}$ as $t\to0$, then it is easy: $$ \sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} =x\left(\sqrt[3]{{1 + 5/x}} - \sqrt {1 - 2/x}\right)=x\left(1 + 5/(3x) - 1 + 1/x+ \hbox{h. o. t.}\right)=8/3+ \hbox{h. o. t.}\to8/3 $$

Intelligenti pauca
  • 50,470
  • 4
  • 42
  • 77
1

We can also solve Using $$\displaystyle \bullet\; (1+x)^{n} = 1+nx+\frac{n(n-1)x^2}{2}+..........+\infty$$

So Here Limit $$\displaystyle \lim_{x\rightarrow \infty}\left[\sqrt[3]{x^3+5x^2}-\sqrt{x^2-2x}\right] = \lim_{x\rightarrow \infty}x\cdot \left[\left(1+\frac{5}{x}\right)^{\frac{1}{3}}-\left(1-\frac{2}{x}\right)^{\frac{1}{2}}\right]$$

So limit $$\displaystyle \lim_{x\rightarrow \infty}x\cdot \left[\left(1+\frac{5}{3x}+\frac{1}{3}\cdot -\frac{2}{3}\cdot \frac{25}{x^2}+........\infty\right)-\left(1-\frac{2}{2x}+\frac{1}{2}\cdot -\frac{1}{2}\cdot \frac{4}{x^2}.......\infty\right)\right]$$

So we get $$\displaystyle \lim_{x\rightarrow \infty}x\cdot \left[\frac{5}{3x}+\frac{1}{x}\right] = \frac{8}{3}$$

Here we Negelect Higher power of $x$ in Denominator, bcz when $\lim_{x\rightarrow \infty}\;,$ Then $\displaystyle \frac{1}{x^n}\rightarrow 0\; \forall n\geq 1$

juantheron
  • 53,015
1

The conjugate expression of a mix of cube and square roots is quite complicated. It is much simpler to use Taylor's formula at order $1$: \begin{align*} \sqrt[3]{x^3+5x}&=x\sqrt[3]{1+\frac5x}=x\biggl(1+\frac5{3x} +o\Bigl(\frac1x\Bigr)\biggr)=x+\frac53+o(1)\\ \sqrt{x^2-2x}&=x\sqrt{1-\frac2x}=x\biggl(1-\frac1x +o\Bigl(\frac1x\Bigr)\biggr)=x-1+o(1) \end{align*} (we may suppose $x>0$), hence: $$\sqrt[3]{x^3+5x}-\sqrt{x^2-2x}=\frac83+o(1)\xrightarrow[x\to+\infty]{}\frac83.$$

Bernard
  • 175,478
0

$$(\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} $$ Use Binomial theorem $$[x^3+5x^2]^{1/3} - [x^2 - 2x]^{1/2}$$ $$[x^3(1+\frac{5x^2}{x^3}]^{1/3}-[x^2(1-\frac{2x}{x^2}]^{1/2}$$ $$\approx x(1+\frac{5}{3x}.....)-x(1-\frac{1}{x}.....)$$ $$\approx [x+\frac{5}{3}]-[x-1]$$

Ganesh
  • 350