Problem using L'Hospiital's Rule (indeterminate form: INF - INF)

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In summary, the conversation is about finding the limit as x approaches infinity of the expression (\sqrt{x^2+x} - x) and the different methods that can be used to solve it, such as rationalizing, factoring, and using L'hôpital's rule. The suggested method is to multiply by (sqrt(x^2+x)+x)/(sqrt(x^2+x)+x) to rationalize the expression and then use the limit rules to simplify it further. The conversation ends with the participants expressing their appreciation for the different methods suggested.
  • #1
SpicyPepper
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Homework Statement



[tex]\stackrel{lim}{x\rightarrow\infty}(\sqrt{x^2+x} - x)[/tex]

I have no idea how to do this. In my book, it says I want to convert [tex]\infty - \infty[/tex] forms into a quotient by getting a common denominator, rationalization or by factoring.
 
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  • #2
I would say you want to multiply by (sqrt(x^2+x)+x)/(sqrt(x^2+x)+x). I.e. rationalize.
 
  • #3
Many, many things would work, really. Multiplying and dividing by 1/x, for example.

Why 1/x? Well, it might be more clear if you think of it as "factoring out an x" -- then you just move the x to be a 1/x in the denominator so that you can use L'hôpitals.
 
  • #4
how about doing this :

y = sqrt(x^2 + x) - x

= sqrt(x ( x + 1) ) - x

= sqrt(x) * sqrt(x+1) - x

The use the product rule for term 1, combined with chain rule.

or
y = sqrt(x^2 + x) - x

= 1/x(sqrt(x^2 + x) - x)
= 1/x *sqrt(x^2 + x) - 1
= sqrt( (1/x)^2 ) * sqrt ( x^2 + x) - 1
= sqrt( (1/x)^2 ( x^2 + x) ) - 1
= sqrt( 1 + 1/x ) - 1

the use l'hopital rule.
 
  • #5
Dick said:
I would say you want to multiply by (sqrt(x^2+x)+x)/(sqrt(x^2+x)+x). I.e. rationalize.
^do that

[tex] (\sqrt{x^2+x}-x).\frac{(\sqrt{x^2+x}+x)}{(\sqrt{x^2+x}+x)} = \frac{1}{\frac{x}{x}\sqrt{1+\frac{1}{x^2}}+1} [/tex]
 
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  • #6
I tried out all your suggestions. This is really cool. Thx everyone.
 

1. What is L'Hospital's Rule?

L'Hospital's Rule is a mathematical rule used to evaluate limits of indeterminate forms, specifically when the limit takes on a form of infinity minus infinity or zero over zero. It is named after the French mathematician Guillaume de l'Hospital.

2. How does L'Hospital's Rule work?

L'Hospital's Rule states that for an indeterminate form of infinity minus infinity or zero over zero, the limit can be evaluated by taking the derivative of both the numerator and denominator and then evaluating the resulting limit.

3. When should L'Hospital's Rule be used?

L'Hospital's Rule should only be used when the limit takes on an indeterminate form of infinity minus infinity or zero over zero. It cannot be used for other types of indeterminate forms such as infinity times zero or infinity to the power of zero.

4. Are there any restrictions when using L'Hospital's Rule?

Yes, there are some restrictions when using L'Hospital's Rule. The rule can only be used if the limit exists, the numerator and denominator are both differentiable, and the limit of the derivative of the denominator is not equal to zero.

5. Can L'Hospital's Rule be used multiple times?

Yes, L'Hospital's Rule can be applied multiple times as long as the resulting limit still takes on an indeterminate form of infinity minus infinity or zero over zero. However, it is important to note that using the rule too many times can lead to incorrect results, so it should be used with caution.

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