- #1

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**help!!! limit**

find the limit as p goes to infinity

a_p = sqrt(p^2+p)-p

really dunno how to solve this... i know the limit is 1/2, but i need to prove that 1/2 is really the limit!1

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- Thread starter mathjojo
- Start date

- #1

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find the limit as p goes to infinity

a_p = sqrt(p^2+p)-p

really dunno how to solve this... i know the limit is 1/2, but i need to prove that 1/2 is really the limit!1

- #2

- 690

- 6

Have you realised the result of each of [tex]a_p=\sqrt{(p^2+p)}-p[/tex]

Plug in the numbers 1 to 15? What do you get? What can you therefore tentatively assume about the limit? Now plug in the numbers from 15 to 30, what do you get? Again, what tentative assumption can you make? now 100 to 150?

As p increases[tex]a_p[/tex] becomes what?

What do you get with 1000? Now with 1001? Now with 1002?

Therefore what's the limit

[tex]\lim_{p\rightarrow\infty}[/tex]

and why?

Plug in the numbers 1 to 15? What do you get? What can you therefore tentatively assume about the limit? Now plug in the numbers from 15 to 30, what do you get? Again, what tentative assumption can you make? now 100 to 150?

As p increases[tex]a_p[/tex] becomes what?

What do you get with 1000? Now with 1001? Now with 1002?

Therefore what's the limit

[tex]\lim_{p\rightarrow\infty}[/tex]

and why?

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- #3

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However, to actually prove it, try the method of "conjugates". What is the conjugate of the function? Multiply and divide by the conjugate. The limit works out after some algebra.

- #4

HallsofIvy

Science Advisor

Homework Helper

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I thought I had already replied to this:

One good way of getting rid of square roots is to multiply numerator and denominator by the "complement"- here [itex]\sqrt{p^2+ p}+ p[/itex].

Also, a good way to handle "[itex]p \leftarrow \infty[/itex] is to divide numerator and denominator by a power of p so that you are left with 1/p which goes to 0.

(I can just imagine standing for your oral defense of your doctoral dissertation giving a "moral justification" of your result!)

One good way of getting rid of square roots is to multiply numerator and denominator by the "complement"- here [itex]\sqrt{p^2+ p}+ p[/itex].

Also, a good way to handle "[itex]p \leftarrow \infty[/itex] is to divide numerator and denominator by a power of p so that you are left with 1/p which goes to 0.

(I can just imagine standing for your oral defense of your doctoral dissertation giving a "moral justification" of your result!)

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