Limits

Limits

Let $y=f\left(x\right)$ be a function of $x.$If $x=a,f\left(x\right)$ takes indeterminate form, the we consider the values of the function which is very near to $a.$ If these values tend to a definite unique number as $x$ tends to $a.$ then the unique number, so obtained is called the limit of $f\left(x\right)$at$x=a$and we were it as $\underset{x\to a}{\lim }f\left(x\right).$

Left Hand and Right Hand limits

If values of the function at the points which are very near to $a$on the left tends to definite unique number as $x$ tends to $a,$then the unique number, so obtained is called the left hand limits of $f\left(x\right)$at$x=a,$we write it as

$f\left(a-0\right)=\underset{x\to a^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(a-h\right)$

Similarly, right hand limit write it as

$f\left(a+0\right)=\underset{x\to a^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(a+h\right)$

Existence of Limit $\underset{x\to a}{\lim }f\left(x\right)$ exists, if

1. $\underset{x\to a^{-}}{\lim }f\left(x\right)$ and $\underset{x\to a^{+}}{\lim }f\left(x\right)$ both exist
2. $\underset{x\to a^{-}}{\lim }f\left(x\right)$=$\underset{x\to a^{+}}{\lim }f\left(x\right)$

Uniqueness of Limit If $\underset{x\to a}{\lim }$$f\left(x\right)$ exists, then it is unique. There cannot be two distinct numbers $l_1$and $l_2$ such that when $x$ tends to $a,$ the function $f\left(x\right)$ tends to both $l_1$and $l_2$.

Fundamental Theorems on limits

If $f\left(x\right)$ and $g\left(x\right)$ functions of $x$such that $\underset{x\to a}{\lim }$$f\left(x\right)$ and $\underset{x\to a}{\lim }g\left(x\right)$both exist. Then$\underset{x\to a}{\lim }g$$\left[f\left(x\right)\pm g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)\pm \underset{x\to a}{\lim }g\left(x\right)$

1. $\underset{x\to a}{\lim }\left[kf\left(x\right)\right]=k\underset{x\to a}{\lim }f\left(x\right),$where $k$ is a fixed real number.
2. $\underset{x\to a}{\lim }\left[f\left(x\right)g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)\underset{x\to a}{\lim }g\left(x\right)$
3. $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to a}{\lim }f\left(x\right)}{\underset{x\to a}{\lim }g\left(x\right)},\underset{x\to a}{\lim }g\left(x\right)\ne 0$
4. $\underset{x\to a}{\lim }{\left[f\left(x\right)\right]}^{g\left(x\right)}{\left[\underset{x\to a}{\lim }f\left(x\right)\right]}^{\underset{x\to a}{\lim }g\left(x\right)}$
5. $\underset{x\to a}{\lim }g\left[f\left(x\right)\right]=g\left[\underset{x\to a}{\lim }f\left(x\right)\right]$
6. $\underset{x\to a}{\lim }\log f\left(x\right)=\log \left[\underset{x\to a}{\lim }f\left(x\right)\right]$, provided;$\underset{x\to a}{\lim }f\left(x\right)>0.$
7. $\underset{x\to a}{\lim }{e}^{f\left(x\right)}=e^{\underset{x\to a}{\lim }f\left(x\right)}$
8. If $f\left(x\right)\le g\left(x\right)$ for every $x$ excluding $a,$ then;$\underset{x\to a}{\lim }f\left(x\right)$$\le \underset{x\to a}{\lim }g\left(x\right).$
9. $\underset{x\to a}{\lim }f\left|f\left(x\right)\right|=\left|\underset{x\to a}{\lim }f\left(x\right)\right|$
10. If $\underset{x\to a}{\lim }f\left(x\right)=+\infty or-$then $\underset{x\to a}{\lim }\frac{1}{f\left(x\right)}=0$

Important Results on Limits

A. Algebraic Limits

1. $\underset{x\to a}{\lim }\frac{x^n-a^n}{x-a}=n{a}^{n-1}$
2. $\underset{x\to a}{\lim }\frac{{\left(1+x\right)}^n-1}{x}=n$

B. Trigonometric Limits

1. $\underset{x\to a}{\lim }\frac{\sin x}{x}=1=\underset{x\to a}{\lim }\frac{x}{\sin x}$
2. $\underset{x\to a}{\lim }\frac{\tan x}{x}=1=\underset{x\to a}{\lim }\frac{x}{\tan x}$
3. $\underset{x\to a}{\lim }\frac{{\sin }^{-1}x}{x}=1=\underset{x\to a}{\lim }\frac{x}{{\sin }^{-1}x}$
4. $\underset{x\to a}{\lim }\frac{{\tan }^{-1}x}{x}=1=\underset{x\to a}{\lim }\frac{x}{{\tan }^{-1}x}$
5. $\underset{x\to a}{\lim }\frac{\sin x{}^{\text{o}}}{x}=\frac{\pi }{180{}^{\text{o}}}$
6. $\underset{x\to a}{\lim }\cos x=1$
7. $\underset{x\to a}{\lim }\frac{\sin \left(x-a\right)}{x-a}=1$
8. $\underset{x\to a}{\lim }\frac{\tan \left(x-a\right)}{x-a}=1$
9. $\underset{x\to a}{\lim }{\sin }^{-1}x={\sin }^{-1}a,\left|a\right|\le 1$
10. $\underset{x\to a}{\lim }{\cos }^{-1}x={\sin }^{-1}a,\left|a\right|\le 1$
11. $\underset{x\to a}{\lim }{\tan }^{-1}x={\tan }^{-1}a,-\infty <a<\infty$
12. $\underset{x\to a}{\lim }\frac{\sin x}{x}=\underset{x\to a}{\lim }\frac{\cos x}{x}=0$
13. $\underset{x\to a}{\lim }\frac{\sin \frac{1}{2}}{\frac{1}{x}}=1$
14. $\underset{x\to a}{\lim }\frac{1-\cos x}{x}=a$

C. Exponential Limits

We use the series $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}=+.........\infty$

1. $\underset{x\to a}{\lim }\frac{e^x-1}{x}$
2. $\underset{x\to a}{\lim }\frac{a^x-1}{x}={\log }_{e}a,a>0$
3. $\underset{x\to a}{\lim }\frac{{e^{\lambda }}^x-1}{x}=\lambda ,$ where $\left(\lambda \ne 0\right).$

D. Logarithmic Limits

We have series $\log \left(1+x\right)=x-\frac{x^2}{2}+\frac{x^2}{3}-....\infty$

Where, $-1<x\le 1$ and expansion is true only, if base is e.

1. $\underset{x\to a}{\lim }\frac{{\log }_e\left(1+x\right)}{x}=1$
2. $\underset{x\to a}{\lim }{\log }_{e}x=1$
3. $\underset{x\to 0}{\lim }\frac{{\log }_e\left(1-x\right)}{x}=1$
4. $\underset{x\to 0}{\lim }\frac{{\log }_e\left(1-x\right)}{x}={\log }_{a}e,a>0,\ne 1$
5. If $\underset{x\to a}{\lim }f\left(x\right)$exists and positive, then

$\underset{x\to a}{\lim }{\left[f\left(x\right)\right]}^{\varphi \left(x\right)}=e^{\underset{x\to a}{\lim }\varphi (x)logf\left(x\right)}$

E. Based on the From $1^{\infty }$

To evaluate the exponential form $1^{\infty }$, we use following results.

If $\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }g\left(x\right)=0,$

Then, $\underset{x\to a}{\lim }{\left\{1+f\left(x\right)\right\}}^{1/g\left(x\right)}{e}^{\underset{x\to a}{\lim }g\frac{f\left(x\right)}{g\left(x\right)}}$

Or If $\underset{x\to a}{\lim }f\left(x\right)=1$and $\underset{x\to a}{\lim }g\left(x\right)=\infty$,

Then , $\underset{x\to a}{\lim }{\left\{f\left(x\right)\right\}}^{g\left(x\right)}=\underset{x\to a}{\lim }\left\{1+f\left(x\right)-1\right\}g\left(x\right)=e^{\underset{x\to a}{\lim }\left\{f\left(x\right)-1\right\}g\left(x\right)}$

1. $\underset{x\to 0}{\lim }{\left(1+x\right)}^{\frac{1}{x}}=e$
2. $\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x=e$
3. $\underset{x\to 0}{\lim }{\left(1+\lambda x\right)}^{\frac{1}{x}}=e^{\lambda }$
4. $\underset{x\to 0}{\lim }{\left(1+\frac{\lambda }{x}\right)}^x=e^{\lambda }$
5. $\underset{x\to \infty }{\lim }{a}^x=\{\begin{array}{c}0,0\le a<1\\ 1,a=1\\ \infty a>1\\ Doesnotexista<0\end{array}$

Methods of Evaluating Limits

1. Determinate Forms (Limits by Direct substitution)

To find $\underset{x\to a}{\lim }f\left(x\right)$, we substitute $x=a$ in the function. If the value comes out to be a definite value, then it is the limit. i.e. $\underset{x\to a}{\lim }f\left(x\right)$=$f\left(a\right)$ Provided it exists.

2. Indeterminate Forms

If direct substitution of $x=a$ while evaluating $\underset{x\to a}{\lim }f\left(x\right)$ leads to one of the following form

$\frac{0}{0};\frac{\infty }{\infty };\infty -\infty ;0\times \infty ;1^{\infty },0$ and $\infty {}^{\text{o}}$

Then, It is called indeterminate form. These limits can be determined by using $L^{\prime }$ Hospital's rule or some other method given below.

3. Limits by Factorisation If $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ attains $\frac{0}{0}$ form, then $x-a$ must be a factor of numerator and denominator which can be cancelled out.

If $n\in Q,$ then $\underset{x\to a}{\lim }\frac{x^n-a^n}{x-a}=n{a}^{n-1}.$

4. Limits by substitution

In order to evaluate$\underset{x\to a}{\lim }f\left(x\right),$we may substitute $x=a+h\left(orx=a-h\right),$so that as$x\to a$it implies that $x\to 0.$ Thus, $\underset{h\to 0}{\lim }f\left(a\pm h\right)$

5. Limits of Functions as $x\to \infty$

If $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is form $\frac{\infty }{\infty }$ and $f\left(x\right)$and $g\left(x\right)$ are both polynomial of $x.$ Then, we divide numerator and denominator by the highest power of $x$and put 0 for $\frac{1}{x}.$

If $m$and$n$ are positive integers and $a_0,b_0\ne 0$are

Non-zero real numbers, then

$\underset{x\to a}{\lim }\frac{a_0{x}^m+a_1{x}^{m-1}+.....+a_{m-1}x+a_m}{b_0{x}^n+b_1{x}^{n-1}+.....+b_{n-1}x+b_n}$

$=\{\begin{array}{c}\frac{a_0}{b_0}ifm=n\\ 0,ifm=<n\\ \infty ,ifm>n,a_0{b}_0<0.\\ -\infty mifm>n,a_0{b}_0<0.\end{array}$

L'Hospital's Rule

If $f\left(x\right)$ and $g\left(x\right)$ be two functions of $x$ such that

1. $\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }g\left(x\right)=0$
2. Both are continuous $x=a.$
3. Both are differentiable at $x=a.$
4. $f^{\prime }\left(x\right)$and $g^{\prime }\left(x\right)$are continuous at the point $x=a,$ then $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$provided that $g\left(a\right).\ne 0$

Above rule is also applicable, if $\underset{x\to a}{\lim }f\left(x\right)=\infty$and $\underset{x\to a}{\lim }g\left(x\right)=\infty .$

Note. If $\underset{x\to a}{\lim }\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$ assumes the indeterminate form $\frac{0}{0}$ or =$\frac{\infty }{\infty }$$f^{\prime }\left(x\right),g\left(x\right)$satisfy all the condition embedded in $L^{\prime }$ Hospital's rule .We can repeat the application of this rule on of this rule on $\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$ to get $\underset{x\to \infty }{\lim }\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}=$get $\frac{{f^{\prime }}^{\prime }\left(x\right)}{{g^{\prime }}^{\prime }\left(x\right)}.$

Limits Using Expansions

Many limits can be evaluated very easily by applying expansion series. Some of the standard expansions are

1. ${\left(1-x\right)}^n=1+{}^n{C}_{1}x+{}^n{C}_2{x}^2+.....+{}^c{C}_n{x}^n,n\in N,x\in R$
2. ${\left(1+x\right)}^n=1+nx\frac{n\left(n-1\right)}{2!}{x}^2+.....\infty ,-1<x<1,n\in Q$
3. $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\infty ,x\in R$
4. $a^x=1x{\log }_{e}a+{\frac{\left(x{\log }_{e}a\right)}{2!}}^2+....\infty ,\in R,a>0,a\ne 1$
5. ${\log }_e\left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+....\infty ,-1<x\le 1$
6. ${\log }_e\left(1-x\right)=-x-\frac{x^2}{2}-\frac{x^2}{3}-.....\infty ,-1\le x\le 1$
7. $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-....\infty ,x\in R$
8. $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...\infty ,x\in R$
9. $\tan x=x+\frac{x^3}{3}+\frac{2}{15}{x}^5+.....$
10. $\tan x=x+\frac{x^3}{3}+\frac{2}{15}{x}^5+.....$
11. ${\tan }^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+......$

Key Points

1. $\underset{x\to 0}{\lim }\frac{1-\cos mx}{1-\cos nx}=\frac{m^2}{n^2}$
2. $\underset{x\to 0}{\lim }\frac{\cos ax-\cos bx}{\cos cx-\cos dx}=\frac{a^2-b^2}{c^2-d^2}$
3. $\underset{x\to 0}{\lim }\frac{{\sin }^{p}mx-\cos nx}{x^2}=\frac{n^2-x^2}{2}$
4. $\underset{x\to 0}{\lim }\frac{{\sin }^{p}mx}{{\left(nx\right)}^p}={\left(\frac{m}{n}\right)}^p$
5. $\underset{x\to 0}{\lim }\frac{{\tan }^{p}mx}{{\tan }^{p}nx}={\left(\frac{m}{n}\right)}^p$
6. $\underset{x\to 0}{\lim }\frac{a^2-a^x}{x^x-a^a}=\frac{1-\log a}{1+\log a}$
7. $\underset{x\to 0}{\lim }\frac{{\left(1+x\right)}^m-1}{{\left(1+x\right)}^n-1}=\frac{m}{n}$
8. $\underset{x\to 0}{\lim }\frac{{\left(1+bx\right)}^m-1}{{\left(1+ax\right)}^n-1}=\frac{mb}{na}$
9. $\underset{x\to 0}{\lim }{\left(1+ax\right)}^{b/x}=\underset{x\to \infty }{\lim }{\left(1+\frac{a}{x}\right)}^{bx}=e^{ab}$
10. $\underset{x\to \infty }{\lim }{\left(x^n+y^n\right)}^{\frac{1}{n}}=y,\left(0<x<y\right)$
11. $\underset{x\to 0}{\lim }{\left(\cos x+a\sin bx\right)}^{1/x}=e^{ab}$
12. $\underset{x\to \infty }{\lim }\frac{x^n}{e^x}=0,\forall n$
13. $\underset{x\to \infty }{\lim }{\left(\cos \frac{x}{m}\right)}^m=1$
14. $\underset{x\to \infty }{\lim }\cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}....\cos \frac{x}{2^n}=\frac{\sin x}{x}$

Sandwich Theorem

Let $f\left(x\right),g\left(x\right)$and $h\left(x\right)$ be real functions such that

$f\left(x\right)\le g\left(x\right)\le h\left(x\right),\forall x$ in the common domain.

If $\underset{x\to a}{\lim }f\left(x\right)=l=\underset{x\to a}{\lim }h\left(x\right),$

Then$\underset{x\to a}{\lim }g\left(x\right)=l$