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BBy Bot
Nov 03'24
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Assume that [math]a \gt 1[/math].

  • lab{5.4.11a} Using the definition of [math]a^x[/math], show that
    [[math]] \lim_{x\goesto\infty} a^x = \infty . [[/math]]
  • lab{5.4.11b} Using the result of \ref{ex5.4.11a}, prove that
    [[math]] \lim_{x\goesto-\infty} a^x = 0 . [[/math]]
  • lab{5.4.11c} Using \ref{ex5.4.11a}, show that
    [[math]] \lim_{x\goesto\infty} \frac{d}{dx} a^x = \infty . [[/math]]
  • lab{5.4.11d} Using \ref{ex5.4.11b}, show that
    [[math]] \lim_{x\goesto-\infty} \frac{d}{dx} a^x = 0 . [[/math]]
  • What do \ref{ex5.4.11a}, \ref{ex5.4.11b}, \ref{ex5.4.11c}, and \ref{ex5.4.11d} say geometrically about the graph of the function [math]a^x[/math]?
BBy Bot
Nov 03'24
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Assume that [math]a_1 \gt a_2 \gt 1[/math].

  • lab{5.4.12a} Using the definition of [math]a^x[/math], show that, if [math]x \gt 0[/math], then [math]{a_1}^x \gt {a_2}^x[/math].
  • Using \ref{ex5.4.12a}, show that, if [math]x \lt 0[/math], then [math]{a_1}^x \lt {a_2}^x[/math].
BBy Bot
Nov 03'24
[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

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