Limit Theorems

Table of Contents

  1. Basic Limit Theorems
  2. An Important Theorem on Limit
  3. Limits of Algebraic Functions
  4. Limits of Trigonometric Functions
  5. Limits of Logarithmic and Exponential Functions

Let $ƒ(x)$ and $g(x)$ be two functions of $x$ such that \[\lim_{x \to a} ƒ(x)=l\] \[\lim_{x \to a} g(x)=m\] then we have the following theorems on limits:

  1. The limit of the sum or difference of the functions $ƒ(x)$ and $g(x)$ is the sum or difference of the limits of the functions i.e. \[\lim_{x \to a} [ƒ(x)±g(x)]=\lim_{x \to a} ƒ(x) ± \lim_{x \to a} g(x)\]\[=l±m\]
  2. The limit of the product of the functions $ƒ(x)$ and $g(x)$ is the product of the limits of the functions i.e. \[\lim_{x \to a} [ƒ(x).g(x)]=\left(\lim_{x \to a} ƒ(x)\right).\left(\lim_{x \to a} g(x) \right)\] \[=l.m\]
  3. The limit of the quotient of the functions $ƒ(x)$ and $g(x)$ is the quotient of the limits of the functions, provided that the limit of the denominator is not zero i.e. \[\lim_{x \to a} \frac{ƒ(x)}{g(x)}=\frac{\lim_{x \to a} ƒ(x)} {\lim_{x \to a} g(x)}\] \[=\frac{l}{m} \text{ } \left(\lim_{x \to a} g(x)=m≠0\right)\]
  4. The limit of the $n^{\text{th}}$ root of a function $ƒ(x)$ is the $n^{\text{th}}$ root of the limit of the function i.e. \[\lim_{x \to a} \sqrt[n]{ƒ(x)}=\sqrt[n]{\lim_{x \to a} ƒ(x)}=\sqrt[n]{l}\]

An important theorem on limit

For all rational values of $n$, \[\lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}\] The proof of this theorem consists of the following three cases:

Case I: When $n$ is a positive integer: \[\frac{x^n-a^n}{x-a}\]\[=x^{n-1}+x^{n-2}.a+x^{n-3}.a^2+…+a^{n-1}\] Now, \[\lim_{x \to a} \frac{x^n-a^n}{x-a}\]\[=\lim_{x \to a} [x^{n-1}+x^{n-2}.a+x^{n-3}.a^2+…+a^{n-1}]\] \[=a^{n-1}+a^{n-1}+a^{n-1}+…+a^{n-1}\] \[=na^{n-1}\]

Case II: When $n$ is a negative integer:

Let $n=-m$ where $m$ is a positive integer. Then, \[\lim_{x \to a} \frac{x^n-a^n}{x-a}=\lim_{x \to a} \frac{x^{-m}-a^{-m}}{x-a}\] \[=\lim_{x \to a} \frac{\frac{1}{x^m}-\frac{1}{a^m}} {x-a}\] \[=\lim_{x \to a} \frac{a^m-x^m}{x^ma^m(x-a)}\] \[=\lim_{x \to a} \left[-\frac{x^m-a^m}{x-a}×\frac{1}{x^ma^m}\right]\] \[=-\left(\lim_{x \to a} \frac{x^m-a^m}{x-a}\right) \left(\lim_{x \to a} \frac{1}{x^ma^m}\right)\] \[=-m.a^{m-1}\frac{1}{a^m.a^m} \text{ (using case I)}\] \[=(-m)a^{(-m)-1}\] \[=na^{n-1}\]

Case III: When $n$ is a rational fraction:

Let $n=\frac{p}{q}$ where $p$ and $q$ are integers and $q≠0$. Then, \[\lim_{x \to a} \frac{x^n-a^n}{x-a}=\lim_{x \to a} \frac{x^{\frac{p}{q}}-a^{\frac{p}{q}}}{x-a}\] \[=\lim_{x \to a} \frac{x^{(\frac{1}{q})p}-a^{(\frac{1}{q})p}}{x-a}\] Put $x^{\frac{1}{q}}=y$ and $a^{\frac{1}{q}}=b$ so that $x=y^q$ and $a=b^q$. When $x\to a$, $y \to b$. Now, \[\lim_{x \to a} \frac{x^n-a^n}{x-a}=\lim_{y \to b} \frac{y^p-b^p}{y^q-b^q}=\lim_{y \to b} \frac{\frac{y^p-b^p}{y-b}} {\frac{y^q-b^q}{y-b}}\] \[=\frac{\lim_{y \to b} \frac{y^p-b^p}{y-b}} {\lim_{y \to b} \frac{y^q-b^q}{y-b}}=\frac{p.b^{p-1}} {q.b^{q-1}}=\frac{p}{q}b^{p-q}\]\[=\frac{p}{q}b^{q(\frac{p}{q}-1)}=\frac{p}{q}(b^q)^{\frac{p}{q}-1}=na^{n-1}\] Therefore, for all rational values of $n$, \[\lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}\]

Limits of Algebraic Functions

Example 1: Find the limiting value of $ƒ(x)=3x^2-5x+6$ when $x\to 2$. \[\lim_{x \to 2} (3x^2-5x+6)=12-10+6=8\] Example 2: Evaluate \[\lim_{x \to 0} \frac{5x^2+3x}{x}\] When $x=0$, the given function takes the indeterminate form $\frac{0}{0}$. But, \[\lim_{x \to a}\frac{5x^2+3x}{x}=\lim_{x \to 0} \frac{x(5x+3)}{x}\] \[=\lim_{x \to 0} (5x+3)=0+3=3\] Example 3: Evaluate \[\lim_{x \to \infty} \frac{3x^2+2x+1}{4x^2+x+5}\] When $x=\infty$, the given function takes the indeterminate form $\frac{\infty}{\infty}$. Therefore, \[\lim_{x \to a} \frac{3+\frac{2}{x}+\frac{1}{x^2}}{4+\frac{1}{x}+\frac{5}{x^2}}=\frac{3+0+0}{4+0+0}=\frac{3}{4}\] Example 4: Evaluate \[\lim_{x \to \infty} (\sqrt{x+a}-\sqrt{x})\] When $x=\infty$, the given function takes the indeterminate form $\infty-\infty$. But, \[\lim_{x \to \infty} (\sqrt{x+a}-\sqrt{x})\]\[=\lim_{x \to \infty} \frac{(\sqrt{x+a}-\sqrt{x})(\sqrt{x+a}+\sqrt{x})}{(\sqrt{x+a}+\sqrt{x})}\] \[=\lim_{x \to \infty} \frac{x+a-x}{\sqrt{x+a}+\sqrt{x}}=\lim_{x \to \infty} \frac{a}{\sqrt{x+a}+\sqrt{x}}\] \[=\frac{a}{\infty+\infty}=\frac{a}{\infty}=0\]

Limits of Trigonometric Functions

\[\text{1. } \lim_{\theta \to 0} \sin\theta = 0\text{ and, 2. } \lim_{\theta \to 0}\cos\theta=1\]

Let $\angle AOB=\theta$. Take any point $P$ on the line $OB$. From $P$ draw $PM$ perpendicular to $OA$.

Limits of trigonometric functions

Then, \[\sin\theta=\frac{MP}{OP}\text{ and }\cos\theta=\frac{OM}{OP}\] When $\theta$ is small, $MP$ will be small and $P$ will be near to $M$. When $\theta$ is small enough, $MP$ will be small enough and $P$ will be very close to $M$. This implies that as $\theta \to 0$, $MP\to 0$ and $OP\to OM$. \[\therefore \lim_{\theta \to 0} \sin\theta=\lim_{\theta \to 0} \frac{MP}{OP}=0\] \[\lim_{\theta \to 0} \cos\theta=\lim_{\theta \to 0} \frac{OM}{OP}=1\] $\text{3. } \lim_{\theta \to \alpha} \sin\theta=\sin\alpha$

Put $\theta=\alpha+h$ so that when $\theta \to \alpha$, $h\to 0$. Now, \[\lim_{\theta \to \alpha} \sin\theta =\lim_{h \to 0} \sin(\alpha+h)\] \[=\lim_{h \to 0} [sin\alpha\cosh+\cos\alpha\sinh]\] \[=\sin\alpha \lim_{h \to 0} \cos h + \cos\alpha\lim_{h \to 0} \sin h\] \[=\sin\alpha.1+\cos\alpha.0\]\[=\sin\alpha\] \[\therefore \lim_{\theta \to \alpha} \sin\theta = \sin\alpha\]

Theorem: \[\lim_{\theta \to 0} \frac{\sin\theta}{\theta}=1\] Where $\theta$ is measured in radian.

Let $AB$ be a diameter of a circle with centre at $O$ and radius $r$. Take any point $P$ on the circle which is very close to $B$ so that $\theta \to 0$. Join $P$ and $B$ and draw $PN$ perpendicular to $OB$.

Limits of trigonometric functions: theorem lim theta tends to 0 sin theta/theta=1

Then, \[∆OPB≤\text{Area of sector OPB}≤∆OPT\] Now, \[∆OPB=\frac{1}{2}OB.PN=\frac{1}{2}r.r\sin\theta\]\[=\frac{1}{2}r^2\sin\theta\] \[\text{Area of sector OPB}=\frac{1}{2}lr\] \[=\frac{1}{2}r.r\theta=\frac{1}{2}r^2 \theta\] \[∆OPT=\frac{1}{2}OP.PT\] \[=\frac{1}{2}r.r\tan\theta=\frac{1}{2}r^2\tan\theta\] \[\therefore \frac{1}{2}r^2\sin\theta≤\frac{1}{2}r^2\theta≤\frac{1}{2}r^2\tan\theta\] \[\sin\theta≤\theta≤\tan\theta\] \[1≤\frac{\theta}{\sin\theta}≤\frac{1}{\cos\theta}\] \[1≥\frac{\sin\theta}{\theta}≥\cos\theta\] Making $\theta \to 0$, \[\lim_{\theta \to 0} 1 ≥ \lim_{\theta \to 0} \frac{\sin\theta}{\theta} ≥ \lim_{\theta \to 0} \cos\theta\] \[1≥\lim_{\theta \to 0} \frac{\sin\theta}{\theta}≥1\] \[1=\lim_{\theta \to 0} \frac{\sin\theta}{\theta}=1\] \[\therefore \lim_{\theta \to 0} \frac{\sin\theta}{\theta}=1\] Also, $\text{1. } \lim_{\theta \to 0} \frac{\theta}{\sin\theta}=1$ \[\lim_{\theta \to 0} \frac{\theta}{\sin\theta}=\lim_{\theta \to 0} \frac{1}{\frac{\sin\theta}{\theta}}=\frac{1}{\lim_{\theta \to 0}\frac{sin\theta}{\theta}}=1\] $\text{2. } \lim_{\theta \to 0}\frac{\tan\theta}{\theta}=1$ \[\lim_{\theta \to 0} \frac{\tan\theta}{\theta}=\lim_{\theta \to 0} \frac{\sin\theta}{\theta\cos\theta}\]\[=\lim_{\theta \to 0} \frac{\sin\theta}{\theta}\lim_{\theta \to 0} \frac{1}{\cos\theta}=1.\frac{1}{\cos 0}=1\]

Limits of Logarithmic and Exponential Functions

According to the definition of $e$, \[e=\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n\] Putting $n=\frac{1}{h}$ so that when $n\to \infty$, $h\to 0$, then \[\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n=\lim_{h \to 0}(1+h)^{\frac{1}{h}}=e\] $\text{1. } \lim_{x \to 0} \frac{\log(1+x)}{x}=1$ \[\lim_{x \to 0} \frac{\log(1+x)}{x}=\lim_{x \to 0}\frac{1}{x}\log(1+x)\] \[=\lim_{x \to 0} \log(1+x)^{\frac{1}{x}}\] \[=\log \left\{\lim_{x \to 0}(1+x)^{\frac{1}{x}}\right\}=\log e=1\] $\text{2 .} \lim_{x \to 0} \frac{e^x-1}{x}=1$

Put $e^x-1=y$, then, $e^x=1+y$ and $x=\log(1+y)$ so that when $x\to 0$, $y\to 0$. \[\lim_{x \to 0}\frac{e^x-1}{x}=\lim_{y \to 0} \frac{y}{\log(1+y)}\] \[=\lim_{y \to 0} \frac{1}{\frac{1}{y}\log(1+y)}=\frac{1}{1}=1\] $\text{3. }\lim_{x \to 0} \frac{a^x-1}{x}=\log a$

Put $a^x-1=y$, then, $a^x=1+y$ which implies $x\log a=\log(1+y)$ and $x=\frac{\log(1+y)}{\log a}$ so that when $x\to 0$, $y\to 0$. \[\lim_{x \to 0} \frac{a^x-1}{x}=\lim_{y \to 0} \frac{y}{\frac{\log(1+y)}{\log a}}\] \[=\log a\lim_{y \to 0} \frac{1}{\frac{1}{y}\log(1+y)}\] \[=\log a.1=\log a\]


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