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: $\lim_{x\to\infty}x^{e^{-x}}=\lim_{x\to\infty}e^{(\ln x)e^{-x}}=1.$