Indeterminate forms

Indeterminate quotients

  • Consider the limit \(\lim_{x\to\infty}\frac{x^3-2x^2+3}{-2x^3+x-2}\).
    • Note that that the nominator \(\to\infty\), and the denominator \(\to-\infty\) as \(x\to\infty\).
    • Therefore, we call this an indeterminate form of type \(\frac{\infty}{\infty}\).
    • Recall that we have calculated this limit with pulling out powers of \(x\): \[ \lim_{x\to\infty}\frac{x^3-2x^2+3}{-2x^3+x-2} =\lim_{x\to\infty}\frac{x^3(1-2x^{-1}+x^{-3})}{x^3(-2+x^{-2}-2x^{-3})}=-\frac12. \]
  • Consider the limit \(\lim_{x\to0}\frac{\sin x}{x}\).
    • Note that both the nominator and the denominator \(\to0\) as \(x\to0\).
    • Therefore, we call this an indeterminate form of type \(\frac00\).
    • Recall that we have calculated this using \((\sin x)'=\cos x\): \[ \lim_{x\to0}\frac{\sin x}{x}=(\sin x)'(x=0)=\cos(x=0)=1. \]

L'Hôpital's rule

  • Note that why we could compute the limits in both cases is that they were of special form.
    • In general, you can calculate limits like this using the following.
    • L'Hôpital's rule. Let \(f\) and \(g\) be differentiable functions defined on an open interval \(I\) containing \(a\). Suppose that \(g'(x)\ne0\) on \(I\) except possibly at \(a\).
    • Suppose that \(\lim_{x\to a}\frac{f(x)}{g(x)}\) is an indeterminate form of type \(\frac00\), or type \(\frac{\infty}{\infty}\).
    • Then we have \[ \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}, \] provided that the limit on the right exists.
    • It is possible to let \(a=\pm\infty\) in this, if we adjust the assumptions accordingly.
    • Exercises. 4.4: 10, 14, 18, 22, 26, 42, 73, 74, 78, 81

Indeterminate products

  • Indeterminate products.
    • Let \(a\) be a number or \(\pm\infty\).
    • Suppose that \(\lim_{x\to a}f(x)=0\), and \(\lim_{x\to a}g(x)=\pm\infty\).
    • In this case, we can't use a limit law to calculate \(\lim_{x\to a}f(x)g(x)\).
    • We call this kind of limit an indeterminate product of type \(0\cdot\infty\).
    • We can calculate it by transforming it to an indeterminate quotient as \[ fg=\frac{f}{1/g}\text{ or }fg=\frac{g}{1/f}. \]
    • Exercises. 4.4: 44, 46, 48

Indeterminate differences

  • Consider \(\lim_{x\to\infty}(\sqrt{x^2-4x}-\sqrt{x^2+x})\).
    • Note that we can't use a limit law, because both terms of the difference \(\to\infty\) as \(x\to\infty\).
    • Therefore, this is called an indeterminate form of type \(\infty-\infty\).
    • How we can deal with these is to transform them to a quotient.
    • In this particular case, we can use an aritmetic rule: \[\begin{multline*} \lim_{x\to\infty}(\sqrt{x^2-4x}-\sqrt{x^2+x}) =\lim_{x\to\infty}\frac{(\sqrt{x^2-4x}-\sqrt{x^2+x})(\sqrt{x^2-4x}+\sqrt{x^2+x})}{\sqrt{x^2-4x}+\sqrt{x^2+x}}\\ =\lim_{x\to\infty}\frac{-5x}{\sqrt{x^2-4x}+\sqrt{x^2+x}}=-\frac52. \end{multline*}\]
    • Exercises. 4.4: 52, 54, 77

Indeterminate powers

  • The limit \(\lim_{x\to a}(f(x))^{g(x)}\) can give rise to several indeterminate forms.
    • If \(f(x)\to0\) and \(g(x)\to0\) as \(x\to a\), then it is of type \(0^0\).
    • If \(f(x)\to\infty\) and \(g(x)\to0\) as \(x\to a\), then it is of type \(\infty^0\).
    • If \(f(x)\to1\) and \(g(x)\to\pm\infty\) as \(x\to a\), then it is of type \(1^\infty\).
    • In all these, we can either take the natural logarithm: \[ \text{letting }y=(f(x))^{g(x)}\text{, we get }\ln y=g(x)\ln f(x), \] or rewrite the function: \[ (f(x))^{g(x)}=e^{g(x)\ln f(x)}. \]
    • Note that in both cases we need to find the limit of the product \(g(x)\ln f(x)\).
    • Exercises. 4.4: 60, 64, 68

Notes on the problems from last time

  • 4.4.77
    • We were studying the family of graphs \(y=f(x)=e^x-cx\).
    • We have gotten that \(f(x)\) has an absolute minimum precisely when \(c>0\), and in that case, that minimum is \(f(x=\ln c)=c-c\ln c\).
    • Therefore, we get two functions \(g(c)=\ln c\) and \(h(c)=c-c\ln c\). To a choice of the constant \(c\), \(g(c)\) assigns the position of the absolute minimum, and \(h(c)\) assigns the value of the absolute minimum.
    • We have \(g'(c)=c^{-1},\) which is \(>0\) when \(c>0\). This shows that as \(c\) increases, the number \(x=\ln c\) at which \(f(x)\) has a minimum absolute value is increasing.
    • We have \[ h'(c)=1-\ln c-1=-\ln c=\begin{cases} >0 & 0<c<1 \\ 0 & c=1 \\ <0 & 1<c, \end{cases} \] which shows that as \(c\) increases, the absolute minimum value \(f(x=\ln c)=c-c\ln c\) first increases, has a maximum \(h(c=1)=1\), and then decreases.
    • You can check these claims on the graphs for \(c=0,\frac14,\frac24,\dotsc,10\) if you click here

Notes on the problems from last time 2

  • 4.4.64
    • We were studying the limit \(\lim_{x\to\infty}x^{e^{-x}}\).
    • We have found out that it is of type \(\infty^0\), therefore we needed to study \(\lim_{x\to\infty}(\ln x)e^{-x}\).
    • We have found that \(\lim_{x\to\infty}(\ln x)e^{-x}=0\).
    • To state the final result using this: \[ \lim_{x\to\infty}x^{e^{-x}}=\lim_{x\to\infty}e^{(\ln x)e^{-x}}=1. \]