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Arranging a fraction to simplify it 
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#1
Mar1512, 09:02 PM

P: 300

1. The problem statement, all variables and given/known data
[itex]\stackrel{lim}{x\rightarrow}∞[/itex] [itex]\sqrt{x^2+x}x[/itex] I'm taking the limit of a rational function as x approaches ∞. However, within the problem, there is an algebraic arrangement i'm having trouble with. How would I get from the first fraction to the second fraction? Applying L'hopitals to: [itex]\frac{x}{\sqrt{x^2+x}+x}[/itex] I get: [itex]\frac{1}{\frac{x+1}{x+\sqrt{x}}+1}[/itex] But don't know how to get to: [itex]\frac{1}{\sqrt{1+\frac{1}{x}}+1}[/itex] ??? And the final solution simplified: 1/2 


#2
Mar1512, 09:32 PM

P: 38

(If not check Example 2: http://tutorial.math.lamar.edu/Class...InfinityI.aspx) What is your highest power of x in the denominator? (remember that [itex]\sqrt{x^2}[/itex] is not x to the power of 2. 


#3
Mar1612, 12:04 AM

P: 300

Hi, yes we learned about the method of factoring or simplifying the fraction w/o applying L'Hopital Rule. The square root of x^2 is lxl. I did notice that I could have applied this method, however, I believe since we are on the chapter about L'Hopitals, my professor wants us to use L'Hopitals rule so that he knows we know how to do it.
Thus, my confusion is not the calculus portion, but the algebra afterwards. Specifically, I don't know how to get from this: [tex]\frac{1}{\frac{x+1}{x+\sqrt{x}}+1}[/tex] to this: [tex]\frac{1}{\sqrt{1+\frac{1}{x}}+1}[/tex] What algebra rules/concepts should I know/apply here? 


#4
Mar1612, 12:32 AM

Mentor
P: 21,311

Arranging a fraction to simplify it
$$d/dx \sqrt{x^2 + x} = \frac{1}{2\sqrt{x^2 + x}}\cdot (2x + 1) $$ So far, so good, but things start to fall apart after this. $$= \frac{x + 1}{x + \sqrt{x}}$$ 1  minor mistake  (1/2)(2x + 1) = x + 1/2, not x + 1 2  serious mistake  √(x^{2} + x) ≠ x + √x !! This mistake indicate that you don't understand the properties of radicals. There is NO property that says √(a + b) = √a + √b. I don't know for a fact that this was your thinking, but it sure seems like to me. In any case, it's much simpler to not use L'Hopital's Rule at all. The expression in the denominator of the original limit is ## \sqrt{x^2 + x} + x## Just factor x out of these two terms, and after some simplification, you can take the limit directly. 


#5
Mar1612, 03:18 AM

P: 53

in you r question [itex]\lim_{x \to +\infty} \sqrt{x^2+x}x[/itex]
i dont think there id any need to apply L'hopitals to the equation at all....... Hint: write [itex]\sqrt{x^2+x}[/itex] as [itex]\sqrt{x^2(1+\frac{1}{x})}[/itex] [tex]\frac{x}{\sqrt{x^2(1+\frac{1}{x})}+x}[/tex] take the [itex]x^2[/itex] out of the root. cancel out the top and bottom [itex]x's[/itex] then apply the fact that [itex]\lim_{x \to +\infty}\frac{1}{x} = 0[/itex] 


#6
Mar2012, 06:44 PM

P: 300

Thanks Mark44!! Appreciate your feedback. Thanks theeverkid, thank you as well.



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