exercise:D2a3518607

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Evaluation of a limit of the form [math]\lim_{x\goesto a} f(x)^{g(x)}[/math] is not obvious if any one of the following three possibilities occurs.

• [math]\lim_{x\goesto a} f(x) = \lim_{x\goesto a} g(x) = 0[/math].
• [math]\lim_{x\goesto a} f(x) = 1[/math] and [math]\lim_{x\goesto a} g(x) = \infty[/math].
• [math]\lim_{x\goesto a} f(x) = \infty[/math] and [math]\lim_{x\goesto a} g(x) = 0[/math].

These three types are usually referred to, respectively, as the indeterminate forms [math]0^0[/math], [math]1^\infty[/math], and [math]\infty^0[/math]. The standard attack, akin to logarithmic differentiation, is the following: Let

[[math]] h(x) = \ln f(x)^{g(x)} = g(x) \ln f(x) = \frac{\ln f(x)}{\frac1{g(x)}} . [[/math]]

One then applies L'H\^opital's Rule to the quotient, thereby hopefully discovering that [math]\lim_{x\goesto a} h(x)[/math] exists and what its value is. If it does exist, it follows by the continuity of the exponential function that

[[math]] e^{\left[ \lim_{x\goesto a} h(x)\right]} = \lim_{x\goesto a} e^{h(x)} . [[/math]]

But, since

[[math]] e^{h(x)} = e^{\ln f(x)^{g(x)}} = f(x)^{g(x)} , [[/math]]

we therefore conclude that

[[math]] \lim_{x\goesto a} f(x)^{g(x)} = e^{\left[ \lim_{x\goesto a} h(x)\right]} , [[/math]]

and the problem is solved. Apply this method to evaluate the following limits.

• [math]\lim_{x\goesto0+} x^x[/math]
• [math]\lim_{x\goesto\infty} x^{\frac1x}[/math]
• [math]\lim_{x\goesto0} (1+2x)^{\frac1x}[/math].