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Please I want to check my answer weither is wrong or right, I found that this limit is equal to 1/3, can you check for me please?

$\lim _{x\to \infty \:}\left(x\left(\sqrt{x^2+1}-\sqrt[3]{x^3+1}\right)\right)$

Without l'hopital, thanks in advance

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  • $\begingroup$ what are we allowed to use? can we use Taylor expansion? $\endgroup$ Commented Oct 15, 2015 at 11:36
  • $\begingroup$ @H.R. No, just algebra. We don't know Taylor expansion, I searched it, I saw it in wiki, but we don't have such thing. Thanks ! $\endgroup$ Commented Oct 15, 2015 at 11:38

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HINT:

Set $1/x=h\implies h\to0^+$

$F=\lim _{x\to \infty \:}\left(x\left(\sqrt{x^2+1}-\sqrt[3]{x^3+1}\right)\right)$ $=\lim_{h\to0^+}\dfrac{(1+h^2)^{1/2}-(1+h^3)^{1/2}}{h^2}$

As lcm$(2,3)=6,$ use $a^6-b^6=(a-b)(a^5+a^4b+\cdots+ab^4+b^5)$ to get

$F=\lim_{h\to0^+}\dfrac{(1+h^2)^3-(1+h^3)^2}{h^2}\cdot\dfrac1{\lim_{h\to0^+}\sum_{r=0}^5((1+h^2)^{1/2})^r\cdot((1+h^3)^{1/3})^{(5-r)}}$

$=\lim_{h\to0^+}\dfrac{1+3h^2+3h^4+h^6-(1+3h^3+h^6)}{h^2}\cdot\dfrac1{\sum_{r=0}^51}=\dfrac36$

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    $\begingroup$ Nice solution. There is a small typo error in second equality. :) $${\left( {1 + {h^3}} \right)^{{1 \over 3}}}$$ $\endgroup$ Commented Oct 15, 2015 at 11:50
  • $\begingroup$ Yeah, but the sequel is true. :) $\endgroup$ Commented Oct 15, 2015 at 11:52
  • $\begingroup$ @lab bhattacharjee Could you explain where ${1/h^2}$ cames from here ? $=\lim_{h\to0^+}\dfrac{(1+h^2)^{1/2}-(1+h^3)^{1/2}}{h^2}$ $\endgroup$ Commented Oct 15, 2015 at 11:58
  • $\begingroup$ @H.R. Please I didn't understand why ${1/h^2}$ here $=\lim_{h\to0^+}\dfrac{(1+h^2)^{1/2}-(1+h^3)^{1/2}}{h^2}$ as he set ${1/x} = h $ $\endgroup$ Commented Oct 15, 2015 at 12:00
  • $\begingroup$ Do some algebraic manipulations! :) $$\eqalign{ & x\left( {\sqrt {{x^2} + 1} - \root 3 \of {{x^3} + 1} } \right) \cr & = {1 \over h}\left( {\sqrt {{1 \over {{h^2}}} + 1} - \root 3 \of {{1 \over {{h^3}}} + 1} } \right) = {1 \over h}\left( {\sqrt {{{1 + {h^2}} \over {{h^2}}}} - \root 3 \of {{{1 + {h^3}} \over {{h^3}}}} } \right) \cr & = {1 \over h}\left( {{1 \over {\left| h \right|}}\sqrt {1 + {h^2}} - {1 \over h}\root 3 \of {1 + {h^3}} } \right) = {1 \over {{h^2}}}\left( {{h \over {\left| h \right|}}\sqrt {1 + {h^2}} - \root 3 \of {1 + {h^3}} } \right) \cr} $$ $\endgroup$ Commented Oct 15, 2015 at 12:05
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By using the definitions of $(\sqrt x)'$and$(\sqrt[3]x)'$at $x=1$

As $y=1/x$, it becomes $$\begin{align} \lim_{y\to0}\frac{(1+y^2)^{1/2}-(1+y^3)^{1/3}}{y^2} &=\lim_{y\to0}\frac{(1+y^2)^{1/2}-1}{y^2}-\lim_{y\to0}y\left(\frac{(1+y^3)^{1/3}-1}{y^3}\right)\\&=1/2-0/3=1/2 \end{align}$$

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  • $\begingroup$ $(\sqrt x)'$and$(\sqrt[3]x)'$at $x=1$ not allowed $\endgroup$ Commented Oct 15, 2015 at 12:34

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