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    Show that \lim_{x \to 0} \left(\dfrac{1}{2}\left(1 + a^x\right)\right)^{\frac{1}{x}} = \sqrt{a}.

    I have no idea how to begin to show this, or the method required.
    Any help would be great.
    • PS Helper
    • Study Helper

    PS Helper
    Study Helper
    This may not be the neatest way to do it, but the way that occurred to me straight away was that it would be handy to work with logs because of the scary 1/x.
    Hence the required limit is e^{lim_{x\rightarrow0}[\frac{1}{h} (log(1+a^h)-log(2))]}. (This works because exp and log are continuous functions, so limits pass between them.)
    The argument of exp here is easy enough to evaluate with L'Hopital's rule (which was the obvious thing to use once we notice the "zero divided by zero" form).

    you can use the fact that, for any expression: b^{x}=e^{xln(b)}

    to turn the above (the entire thing in brackets) into an expression in "e" (looks more complicated, but leads to the answer)

    you then get:

    \displaystyle exp(\frac{1}{x} \times ln(\frac{1+a^{x}}{2}))

    split the ln terms up to get: ln(a)-ln(b), then use the Maclaurin series expansion of the first term ( ln(1+a^{x})), subtract the second (ln(2)).

    all terms drop out as x ->0 exept the term (e to the power of):

    \frac{1}{x}(\frac{1}{2}ln(a) \times x)...


    Logs is the way to go indeed. In fact you can then show that in general,

    \displaystyle \lim_{x\to 0} \bigg(\frac{a_1^x+a_2^x+\cdots a_n^x}{n}\bigg)^{\frac{1}{x}}= \big(a_1a_2\cdots a_n\big)^{\frac{1}{n}}

    (i.e. the geometric mean of those numbers)
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