For what value of a is the following equation true? lim x -> Infinite of ( (x+a)

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Using the identity t = e^(ln t), valid for any t > 0, gives rise to the identity t^s = (e^(ln t))^s = e^(s ln t), valid for any t > 0 and any s. Applying this identity here to t = (x + a)/(x - a) and s = x you find that ((x + a)/(x - a))^x = e^(x ln((x + a)/(x - a))) We focus for the moment on just the x ln((x + a)/(x - a)) in the exponent. It is equal to [ln((x + a)/(x - a)]/[1/x] = [ln((1 + a/x)/(1 - a/x))]/[1/x] and the reason I wrote it this way is that it makes clear that as x goes to infinity, no matter what a is, both the numerator and denominator of this fraction go to 0 (the top [ ] goes to ln((1+0)/(1-0)) = ln(1) = 0 and the bottom clearly goes to 0). So we can use L'Hopital's rule to figure out what the limit of this fraction is as x goes to infinity. The derivative of ln((x+a)/(x-a)) = ln(x + a) - ln(x - a) is equal to 1/(x + a) - 1/(x - a) and the derivative of 1/x is -1/x^2, so by L'Hopital's rule, lim_{x to oo} [ln((1 + a/x)/(1 - a/x))]/[1/x] = lim_{x to oo} [1/(x + a) - 1/(x - a)]/[-1/x^2] To figure out this limit it helps to do some algebra: 1/(x + a) - 1/(x - a) = (x - a)/((x + a)(x - a)) - (x + a)/((x + a)(x - a)) = [(x - a) - (x + a)]/[x^2 - a^2] = -2a/(x^2 - a^2) So lim_{x to oo} [1/(x + a) - 1/(x - a)]/[1/x^2] = lim_{x to oo} 2a x^2/(x^2 - a^2) = lim_{x to oo} [2a x^2 * (1/x^2)]/[(x^2 - a^2) * (1/x^2)] = lim_{x to oo} 2a/(1 - (a^2/x^2)) = 2a/(1 - 0) = 2a. To remind ourselves what we have done, we have used L'Hopital's rule and some algebra to show that the limit as x goes to oo of x ln((x + a)/(x - a)) is 2a. It follows that the limit as x goes to oo of ((x + a)/(x - a))^x = e^(x ln((x + a)/(x - a))) is e^(2a). This will be equal to e = e^1 when 2a = 1, or when a = 1/2. The trick of writing f(x)^g(x) as e^(g(x) ln f(x)) is a very common tool in working with functions of this sort. The point is that it replaces the difficulty of dealing with an exponential function where the base and exponent are both functions of x, to dealing with one where the base is just e. This is much easier to analyze. (The formula for the derivative of f(x)^g(x) is easily found in this way, for example: use the identity and then apply the chain and product rules.)

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