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I was trying all the day to resolve this problem with different method without L'Hopital but I can't do it, I would really like to up my mathematical development but the post doesn't allow me because I have less than 10 of reputation to up a image, so here is a link to one of me best try

$$ \lim_{x\rightarrow 1}\left(\frac{\sqrt[3]{7+x^3}-\sqrt[2]{3+x^2}}{x-1}\right) $$

Robert Howard
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mrpepo877
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  • Why? $\qquad \qquad$ –  Jul 29 '18 at 21:37
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    Regarding your try, it is false in general that $\sqrt{a}+\sqrt{b}=\sqrt{a+b}$. – cansomeonehelpmeout Jul 29 '18 at 21:37
  • it was a test that i did yesterday but i couldn t do that problem, so i am tring to solve now – mrpepo877 Jul 29 '18 at 21:38
  • The factorization $$(a^n-b^n)=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+b^{n-1})$$ often helps in instances with roots. If the roots are the same, you can blindly use this. If the roots are different, you'd choose $n$ so that it is the least common multiple of the roots involved. –  Jul 29 '18 at 21:49
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    Hint: $$\lim_{x\to1}\frac{\root3\of{7+x^3}-2}{x-1}-\lim_{x\to1}\frac{\sqrt{3+x^2}-2}{x-1}.$$ Then use $a^3-b^3=(a-b)(\cdots)$ and $a^2-b^2=(a-b)(\cdots)$. – Jyrki Lahtonen Jul 30 '18 at 05:58

2 Answers2

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Hint: Use the (rather lengthy, but perfectly working) identity $$a-b=\frac{a^6-b^6}{a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5}$$

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$$\lim_{x\to1}\dfrac{(7+x^3)^{1/3}-2}{x-1}=\dfrac{d(7+y^3)^{1/3}}{dy}_{(\text{ at }y=1)}=?$$

lab bhattacharjee
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