Question #9f75a

1 Answer
Nov 24, 2017

The limit is e^3.

Explanation:

The keys to solving this are to take the logarithm of the given function then use Taylor series centered at 0 (Maclaurin series) to determine how the expressions [f(x)] and {f(x)} can be simplified when x is close to zero. Then use Taylor series again to calculate the resulting limit before exponentiating to get the final answer.

Let f(x)=tan(x)/x, let g(x)=([f(x)]+x^2)^{1/{{f(x)}}}, and let
h(x)=ln(g(x))=(ln([f(x)]+x^2))/({f(x)}), where the last equality follows by a property of logarithms.

Now when x is close to zero (within 1 unit of 0 is good enough in this case), then [f(x)]=[tan(x)/x]=1 since 1 < tan(x)/x < 2 when -1 < x < 1 (also see the Taylor series expansions below).

When -1 < x < 1 and x!=0 we also have {f(x)}={tan(x)/x}=tan(x)/x-1=x^2/3+2/15 x^4+cdots. The reason for this follows from the Taylor expansion for f(x)=tan(x), which starts tan(x)=x+x^3/3+2/15 x^5+cdots.

This implies that tan(x)/x=1+x^2/3+2/15 x^4+cdots for -1 < x < 1 when x!=0 (actually it works for -pi/2 < x < pi/2 when x!=0).

Certainly then, when x is sufficiently close to zero it follows that {tan(x)/x}=x^2/3+2/15 x^4+cdots. But this is equal to tan(x)/x-1 for x sufficiently close to zero.

The Taylor series for ln(1+x^2) is ln(1+x^2)=x^2-x^4/2+x^6/3-x^8/4+cdots for -1 < x < 1.

Hence, when x is sufficiently close to zero, h(x)=ln(1+x^2)/(tan(x)/x-1)=(x^2-x^4/2+cdots)/(x^2/3+2/15 x^4+cdots).
As x->0, the dominant terms are those of the lowest powers, so lim_{x->0}h(x)=1/(1/3)=3 (L'Hopital's Rule could also be used twice here).

Therefore, lim_{x->0}g(x)=lim_{x->0}e^{h(x)}=e^{lim_{x->0}h(x)}=e^{3}.