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 September 14th, 2013, 03:50 AM #1 Member   Joined: Sep 2012 Posts: 61 Thanks: 0 Limit with cube and square root How to solve this: lim [cuberoot(7+x^3)-squareroot(3+x^2)]/(x-1) x->1 I don't know what to do when I have cube and square root at the same time.
 September 14th, 2013, 04:41 AM #2 Member   Joined: Apr 2013 Posts: 52 Thanks: 0 Re: Limit with cube and square root Since both the numerator and the denominator equal to zero,when x->1,you can apply the "L'Hospital's Rule",that allows you to evaluate the limit more easily!! :P Are you familiar with this rule,or should I explain it further to you??
 September 14th, 2013, 05:24 AM #3 Member   Joined: Sep 2012 Posts: 61 Thanks: 0 Re: Limit with cube and square root I need to do this without differentiation.
 September 15th, 2013, 12:09 AM #4 Member   Joined: Sep 2012 Posts: 61 Thanks: 0 Re: Limit with cube and square root Any idea, please?
 September 16th, 2013, 08:21 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Limit with cube and square root Assuming the limit exists, $\lim_{x\to1}\,\frac{(x^3\,+\,7)^{\frac13}\,-\,(x^2\,+\,3)^{\frac12}}{x\,-\,1} \\ =\,\lim_{x\to1}\,\frac{(x^3\,+\,7)^{\frac23}\,-\,(x^2\,+\,3)}{(x\,-\,1)$(x^3\,+\,7)^{\frac13}\,+\,(x^2\,+\,3)^{\frac12}$} \\ =\,\lim_{x\to1}\,\frac{(x^3\,+\,7)^{\frac23}\,-\,(x\,+\,1)(x\,-\,1)\,-\,4}{(x\,-\,1)$(x^3\,+\,7)^{\frac13}\,+\,(x^2\,+\,3)^{\frac12}$} \\ =\,\lim_{x\to1}\,\frac{$(x^3\,+\,7)^{\frac13}\,-\,2$$(x^3\,+\,7)^{\frac13}\,+\,2$}{(x\,-\,1)$(x^3\,+\,7)^{\frac13}\,+\,(x^2\,+\,3)^{\frac12}$}\,-\,\lim_{x\to1}\,\frac{x\,+\,1}{(x^3\,+\,7)^{\frac1 3}\,+\,(x^2\,+\,3)^{\frac12}} \\ =\,\lim_{x\to1}\,$\frac{\[(x^3\,+\,7)^{\frac13}\,-\,2$$(x^3\,+\,7)^{\frac13}\,+\,2$}{(x\,-\,1)$(x^3\,+\,7)^{\frac13}\,+\,(x^2\,+\,3)^{\frac12}$}\]\,-\,\frac12 \\ =\,\lim_{x\to1}\,\frac{(x^3\,+\,7)^{\frac13}\,-\,2}{x\,-\,1}\,\cdot\,1\,-\,\frac12 \\ \lim_{x\to1}\,\frac{(x^3\,+\,7)^{\frac13}\,-\,2}{x\,-\,1}\,=\,\lim_{x\to1}\,\frac{x^3\,-\,1\,-\,2(x^3\,+\,7)^{\frac13}((x^3\,+\,7)^{\frac13}\,-\,2)}{(x\,-\,1)$(x^3\,+\,7)^{\frac23}\,+\,4)$} \\ \lim_{x\to1}\,\frac{(x^3\,+\,7)^{\frac13}\,-\,2}{x\,-\,1}\,=\,L \\ L\,=\,\frac38\,-\,\frac12L\,\Rightarrow\,L\,=\,\frac14 \\ \lim_{x\to1}\,\frac{(x^3\,+\,7)^{\frac13}\,-\,(x^2\,+\,3)^{\frac12}}{x\,-\,1}\,=\,\frac14\,-\,\frac12\,=\,-\frac14$ Thanks from ocajocaja
 September 16th, 2013, 10:52 AM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Limit with cube and square root I think the limit must exist. Here's why: $\text{Let }L\,=\,\infty\text{ then }\infty\,=\,\frac38\,-\,\frac12\infty\,\Rightarrow\,\infty\,=\,-\infty,\;\text{ a contradiction.}$ Reversing the sign of L gives the same result. Hence the limit exists and is equal to $-\frac14$
September 16th, 2013, 11:26 AM   #7
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Re: Limit with cube and square root

I didn't understand the part I attached. Could you explain more deeply what you did, and with what purpose?
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 September 16th, 2013, 11:42 AM #8 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Limit with cube and square root $\lim_{x\to1}\,\frac{(x^3\,+\,7)^{\frac13}\,-\,2}{x\,-\,1}\,=\,\lim_{x\to1}\,\frac{x^3\,-\,1\,-\,2(x^3\,+\,7)^{\frac13}((x^3\,+\,7)^{\frac13}\,-\,2)}{(x\,-\,1)$(x^3\,+\,7)^{\frac23}\,+\,4)$} \\ =\,\lim_{x\to1}\,\frac{x^3\,-\,1}{(x\,-\,1)$(x^3\,+\,7)^{\frac23}\,+\,4)$}\,-\,\frac{2(x^3\,+\,7)^{\frac13}((x^3\,+\,7)^{\frac1 3}\,-\,2)}{(x\,-\,1)$(x^3\,+\,7)^{\frac23}\,+\,4)$} \\ =\,\frac38\,-\,\frac12\,\cdot\,\lim_{x\to1}\,\frac{(x^3\,+\,7)^ {\frac13}\,-\,2}{x\,-\,1} \\ \lim_{x\to1}\,\frac{(x^3\,+\,7)^{\frac13}\,-\,2}{x\,-\,1}\,=\,L \\ L\,=\,\frac38\,-\,\frac12L\,\Rightarrow\,L\,=\,\frac14 \\ \lim_{x\to1}\,\frac{(x^3\,+\,7)^{\frac13}\,-\,(x^2\,+\,3)^{\frac12}}{x\,-\,1}\,=\,\frac14\,-\,\frac12\,=\,-\frac14$ Ok?
 September 17th, 2013, 07:22 AM #9 Member   Joined: Sep 2012 Posts: 61 Thanks: 0 Re: Limit with cube and square root It's okay, thanks.

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# how to solve limits involving cube roots

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