Solving the limit of a function with no calculator

APcalchelp
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The Limx-->infinity of the function Square rt(5x^4+2x) / (x^2)

Note* The square root is only in the numerator for the expression (5x^4+2x)



I have tried multiplying by the complex conjugate but I am not sure if this is the correct way to do it.

I did sqrt(5x^4-2x) over itself and got sqrt 25x^8-4x / x^2

I had that as my answer, but my teacher told me this was incorrect. If anyone could help and explain that would be great! Thanks a lot!
 
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APcalchelp said:
The Limx-->infinity of the function Square rt(5x^4+2x) / (x^2)

Note* The square root is only in the numerator for the expression (5x^4+2x)



I have tried multiplying by the complex conjugate but I am not sure if this is the correct way to do it.

I did sqrt(5x^4-2x) over itself and got sqrt 25x^8-4x / x^2

I had that as my answer, but my teacher told me this was incorrect. If anyone could help and explain that would be great! Thanks a lot!

Factor x4 out of both terms inside the radical, and bring this out as x2. You should be able to evaluate the limit then.
 
Mark44 said:
Factor x4 out of both terms inside the radical, and bring this out as x2. You should be able to evaluate the limit then.

I'm not clear on what you mean by that. I understand factoring out the x^4 from the 5x^4 but not the 2x.
 
APcalchelp said:
I'm not clear on what you mean by that. I understand factoring out the x^4 from the 5x^4 but not the 2x.

If you factor out x^4 from 2x, you are left with 2/x^3.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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