How to Use L'Hôpital's Rule to Find A in a Limit Equation

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Homework Statement



Find A so that lim (x+A/x-2A)^x =5
x-> 00

Homework Equations





The Attempt at a Solution


got to use lopihtals rule.
 
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Hi gayani! Welcome to PF! :smile:

(have an infinity: ∞ and try using the X2 tag just above the Reply box :wink:)
gayani said:
Find A so that lim (x+A/x-2A)^x =5
x-> 00
…
got to use lopihtals rule.

hmm … I can't see how to use l'Hôpital's rule :redface: …

but if you write the LHS as (1 + 3A/(x - 2A))x, and fiddle about with it, you should get something you recognise. :smile:
 
Gayani,
Just in case you don't follow what Tiny-tim did, here it is.

You wrote the rational expression as x + A/x - 2A. What I'm pretty sure you meant (and the way Tiny-tim interpreted it) was (x + A)/(x - 2A). What you wrote would be correctly interpreted as the sum and difference of three expressions-- x, A/x, and 2/A, not the quotient of x + A and x - 2A.

What Tiny-tim did was divide x + A by x - 2A, so that (x + A)/(x - 2A) = 1 + 3A/(x - 2A).
 
Take the natural log of both sides and then take the limit of the natual log of the left side

\lim_{x\rightarrow\infty} ln\left(1 + \frac{3A} {x-2A}\right)^x= ln (5)

You'll have to do some rearranging in order to use l'Hôpital's rule and "solve" for A.
 
Last edited:

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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