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September 14th, 2013, 03:50 AM   #1
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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.
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September 14th, 2013, 04:41 AM   #2
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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??
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September 14th, 2013, 05:24 AM   #3
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Re: Limit with cube and square root

I need to do this without differentiation.
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September 15th, 2013, 12:09 AM   #4
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Re: Limit with cube and square root

Any idea, please?
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September 16th, 2013, 08:21 AM   #5
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Re: Limit with cube and square root

Assuming the limit exists,

Thanks from ocajocaja
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September 16th, 2013, 10:52 AM   #6
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Re: Limit with cube and square root

I think the limit must exist. Here's why:



Reversing the sign of L gives the same result. Hence the limit exists and is equal to

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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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File Type: gif untitled.gif (8.4 KB, 185 views)
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September 16th, 2013, 11:42 AM   #8
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Re: Limit with cube and square root



Ok?
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September 17th, 2013, 07:22 AM   #9
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Re: Limit with cube and square root

It's okay, thanks.
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