Limits

Limits & Continuity

1. Limit of a function

Let f(x) be defined for all values of x near x = a with the possible exception of x=a itself. If the value of f(x) gets arbitrary close (very close) to L as x get arbitrary close to ' a' we write : 

 $\lim_{x\rightarrow a}f\left(x\right)\ =\ L$  
It is read as " The limit of f(x) as x approaches a is equal to L.'

Note : 
A limit has A unique value.
The limit of a function as x approaches a (
 $\lim_{x\rightarrow a}f\left(x\right)\$  )  will not always be found by subbing a into f(x) [f(a)] .

2) Right and Left hand limits

No restriction is made as to how x should approach 'a' . It can approach 'a' either from the left or from the right. We indicate these 2 approaches by writing  $x\rightarrow a^-$  and  $x\rightarrow a^+$  respectively. 

If  $\lim_{x\rightarrow a^-}f\left(x\right)\ =L_1\ and\ \lim_{x\rightarrow a^+}f\left(x\right)\ =\ L_2$  , then L1 and L2 are respectively the left-hand and right-hand limits of f(x) as  $x\rightarrow a$  .

We say that  $\lim_{x\rightarrow a}f\left(x\right)=\ L\$  exist if and only if  $\lim_{x\rightarrow a^-}f\left(x\right)\ =\ \lim_{x\rightarrow a^+}f\left(x\right)\ =\ L$  . 
(The right hand and left hand limits must be the same. )

E.g : If 
 $\lim_{x\rightarrow3^-}x^2=9$     $\lim_{x\rightarrow3^+}x^{2\ }=9$  
 $\therefore\lim_{x\rightarrow3}x^2=9$  ( IT EXISTS) 

MODULUS WITH LIMITS

3) Theorems of Limits

• (a) If k is a constant ,  $\lim_{x\rightarrow a}k\ =\ k$  

• (b)  $\lim_{x\rightarrow a}k\left(f\left(x\right)\right)\ =\ k\ \times\lim_{x\rightarrow a}f\left(x\right)\$  

• (c)  $\lim_{x\rightarrow a}\left[f\left(x\right)+g\left(x\right)\right]=\lim_{x\rightarrow a}f\left(x\right)+\lim_{x\rightarrow a}g\left(x\right)$  

•  $\left(d\right)\ \lim_{x\rightarrow a}\left[f\left(x\right)-g\left(x\right)\right]=\lim_{x\rightarrow a}f\left(x\right)\ -\lim_{x\rightarrow a}g\left(x\right)$  

• (e)  $\lim_{x\rightarrow a}\left[f\left(x\right)\ \times g\left(x\right)\right]=\lim_{x\rightarrow a}f\left(x\right)\ \times\lim_{x\rightarrow a}g\left(x\right)$  

• (f)  $\lim_{x\rightarrow a}\ \frac{f\left(x\right)}{g\left(x\right)}=\frac{\lim_{x\rightarrow a}f\left(x\right)}{\lim_{x\rightarrow a}g\left(x\right)}\$  

• (g)  $\lim_{x\rightarrow a}\left[f\left(x\right)\right]^n=\left[\lim_{x\rightarrow a}f\left(x\right)\right]^n$  (assuming that the nth power is defined.

continued

•  $\lim_{x\rightarrow a}\ ^n\sqrt{f\left(x\right)}=^n\sqrt{\lim_{x\rightarrow a}f\left(x\right)}$  (Assuming that ythe nth root is defined. )

•  $Note\ :$  

•  $\lim_{x\rightarrow a}x\ =a$  

•  $\lim_{x\rightarrow0}\ \frac{\sin x}{x}=1$   ( $\lim_{x\rightarrow0}\frac{\sin ax}{ax}$   Also counts )

•  $\lim_{x\rightarrow0}\ \frac{1-\cos x}{x}=0$  

•  $\lim_{x\rightarrow+\infty}\left(1+\ \frac{1}{x}\right)^x=e$  

4) X--> Infinity

•  $\frac{\infty}{\infty}=$  indeterminate 

•  $\frac{Z^+}{\infty}=0$  **

•  $\frac{\infty}{Z^+}=\infty$  

Examples

1.  $\lim_{x\rightarrow-3}\ \frac{x^{2\ }-9}{x+3}$                           2.  $\lim_{x\rightarrow1}\ \frac{x^2+x-2}{x-1}$  
 $=\lim_{x\rightarrow-3}\ \frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)}$             $=\lim_{x\rightarrow1}\ \frac{\left(x-1\right)\left(x+2\right)}{x-1}$  
 $=\lim_{x\rightarrow-3}\ \left(x-3\right)$                            $=\lim_{x\rightarrow1}\left(x+2\right)$                                         
 $=\lim_{x\rightarrow-3}x\ -\lim_{x\rightarrow-3}3$                      $=\lim_{x\rightarrow1}x\ +\lim_{x\rightarrow1}2$  
 $=\left(-3\right)-3$                                      = 1 + 2
 $=-6$                                                 = 3

3.  $\lim_{x\rightarrow10}\ \frac{x^{2\ }-19x\ +90}{3x^{2\ }-30x}$                4.  $\lim_{x\rightarrow\infty}\ \frac{x+3}{x+1}$  

 $=\lim_{x\rightarrow10}\ \frac{\left(x-9\right)\left(x-10\right)}{3x\left(x-10\right)}$                 $=\lim_{x\rightarrow\infty}\ \frac{x\left(1+\frac{3}{x}\right)}{x\left(1+\frac{1}{x}\right)}$  
 $=\lim_{x\rightarrow10}\ \frac{x-9}{3x}$                                    $=\lim_{x\rightarrow\infty}\ \frac{1+\frac{3}{x}}{1+\ \frac{1}{x}}$  
 $=\frac{\lim_{x\rightarrow10\ }x\ -\lim_{x\rightarrow10}\ 9}{3\lim_{x\rightarrow10}\ x}$            $=\frac{1+0}{1+0}$  
 $=\frac{10-9}{3\left(10\right)}$                                           = 1
 $=\frac{1}{30}$  

5.  $\lim_{x\rightarrow\infty}\ \frac{2x^2-3x+5}{9x^2+x-1}$               6.  $\lim_{x\rightarrow\infty}\ \frac{x^3+x+2}{x^2+1}$  

 $=\lim_{x\rightarrow\infty}\ \frac{x^2\left(2-\frac{3}{x}+\frac{5}{x^2}\right)}{x^2\left(9+\frac{1}{x}-\frac{1}{x^2}\right)}$              $=\lim_{x\rightarrow\infty}\ \frac{x^3\left(1+\frac{1}{x^2}+\frac{2}{x^3}\right)}{x^2\left(1+\frac{1}{x^2}\right)}$  
 $=\lim_{x\rightarrow\infty}\ \frac{2-\frac{3}{x}+\frac{5}{x^2}}{9+\frac{1}{x}-\frac{1}{x^2}}$                      $=\lim_{x\rightarrow\infty}\ \frac{x\left(1+\frac{1}{x^2}+\frac{2}{x^3}\right)}{\left(1+\frac{1}{x^2}\right)}$         
 $=\ \frac{2-0+0}{9+0-0}$                                 $=\lim_{x\rightarrow\infty}\ \frac{\infty\left(1+0+0\right)}{1+0}$  
 $=\frac{2}{9}$                                                 $=\infty$  

7.  $\lim_{x\rightarrow0}\ \frac{\sin6x}{x}$              8.  $\lim_{x\rightarrow0}\ \frac{\sin x}{\sin\ 3x}$  

 $=\lim_{x\rightarrow0}\ 6\ \times\frac{\sin6x}{6x}$        $=\lim_{x\rightarrow0}\ \frac{x\ \times\frac{\sin\ x}{x}}{3x\ \times\frac{\sin\ 3x}{3x}}$    
 $=6\ \lim_{x\rightarrow0}\ \frac{\sin\ 6x}{6x}$              $=\lim_{x\rightarrow0}\ \frac{\frac{\sin\ x}{x}}{3\ \left(\frac{\sin\ 3x}{3}\right)}$  
 $let\ u\ =6x$                        $=\frac{1}{3}\lim_{x\rightarrow0}\ \frac{\frac{\sin x}{x}}{\frac{\sin3x}{3x}}$  
 $=6\lim_{u\rightarrow0}\ \frac{\sin\ u}{u}$              $=\frac{1}{3}\times\frac{1}{1}$   
 $=6\left(1\right)$                           $=\frac{1}{3}$  
 $=6$  

5) L'Hopital's  Rule

Limits of the form   $\lim\ \frac{f\left(x\right)}{g\left(x\right)}$   can be evaluated by this rule (if they exist) in the indeterminate cases where f(x) and g(x) both approach 0 or both approach +inanity or - infinity . The rule states :

 $\lim\ \frac{f\left(x\right)}{g\left(x\right)}=\lim\ \frac{f^1\left(x\right)}{g^1\left(x\right)},$  
where   $f^1\left(x\right)$   and  $g^1\left(x\right)$   are the first derivatives of f(x) and g(x) respectively, with respect to x. 

 $E.g\ \lim_{x\rightarrow3}\ \frac{x-3}{x^2-9}$        $E.g\ 2\ \lim_{x\rightarrow\infty}\ \frac{3x+2}{5x-7}$  
 $=\lim_{x\rightarrow3}\ \frac{1}{2x}$                              $=\lim_{x\rightarrow\infty}\ \frac{3}{5}$  
 $=\frac{1}{6}$                                         $=\frac{3}{5}$  

5. Continuity

A function f(x) is said to be continuous at x=a if  $\lim_{x\rightarrow a}f\left(x\right)=f\left(a\right)$  . 3 conditions must satisfy. 

1.  $\lim_{x\rightarrow a}\ f\left(x\right)\ =\ L\ must\ exist$  
2. f (a) must exist 
3. f (a) = L 

E.g Consider f(x) =  $x^2$  ,x  $\in$  R
 $\lim_{x\rightarrow2}f\left(x\right)\ =4$  
 $f\left(2\right)=4$  
 $\therefore f\left(x\right)\ is\ continuous\ at\ x=2$  

Types of Discontinuity

If a function is not continuous , it is said to be discontinuous at a point or points.

There types of discontinuity are :

1. Removable or Point

2.Asymptotic

3. Jump

1. Removable or Point Discontinuity

Consider f(x) = $\frac{x^3-8}{x-2},\ x\ne2$  

 $\lim_{x\rightarrow2}\ f\left(x\right)\ =\ \lim_{x\rightarrow2}\ \frac{x^3-8}{x-2}$  
 $=\lim_{x\rightarrow2}\ \frac{\left(x-2\right)\left(x^2+2x+4\right)}{x-2}$  
 $=\lim_{x\rightarrow2}\ x^{2\ }+2x+4$  
 $=\left(2\right)^2+2\left(2\right)+4$  
 $=12$  
f(2) is undefined .
f(x) has a discontinuity at x = 2. It is called a removable or point discontinuity since by extending the definition of f(x) to include a function value of 12 at x = 2, the resulting function g(x) would be continuous at x = 2 , ie
 $g\left(x\right)\ =\left\{\frac{\frac{x^3-8}{x-2},\ x\ne2}{12\ ,\ x=2}\right\}$  
A function has a removable discontinuity at x=a if  $\lim_{x\rightarrow a}\ f\left(x\right)\ =L$  exists but either f(a) does not exist or f(a) s not equal to L. 

2. Asymptotic Discontinuity

Consider g(x) =  $\frac{1}{x^2}$    . This function has a discontinuity at x=0. It is an example of an asymptotic (or infinite) discontinuity.  

e.g

h(x) =  $\frac{2x-4}{x^2-7x+10};\ x\ne2\ ,\ x\ne5$  

 $=\frac{2\left(x-2\right)}{\left(x-2\right)\left(x-5\right)};\ x\ne2,x\ne5$  
 $=\frac{2}{x-5},x\ne5$  

The function has 2 discontinuities - one at x = 2 and the other at x = 5.
 The discontinuity at x = 2 is removable. 
The discontinuity at x = 5 is asymptotic. 

 3. Jump Discontinuity 

 E.g f(x) =  $\frac{\left|x\right|}{x},\ x\ne0$  

 $\lim_{x\rightarrow0^-}\ \frac{\left|x\right|}{x}$                          $\lim_{x\rightarrow0^+}\ \frac{\left|x\right|}{x}$  
 $=\lim_{x\rightarrow0^-}\ \frac{-x}{x}$                     $=\lim_{x\rightarrow0^+\ }\ \frac{x}{x}$  
 $=\lim_{x\rightarrow0^-}\left(-1\right)$                   $=\lim_{x\rightarrow0^+}\ \ \left(1\right)$                                                                                                  $=1$  
 $=-1$                           
 $\therefore\lim_{x\rightarrow0}\ f\left(x\right)\$  does not exist .The function is not continuous at x = 0. We can also come to the same conclusion since f(0) is undefined . The discontinuity at x=0 is called a 'jump' discontinuity. A function has a jump discontinuity at x=a if  $\lim_{x\rightarrow a^+}f\left(x\right)\ \&\lim_{x\rightarrow a^-}\ f\left(x\right)$  both exist but are not equal to each other.

6. Right -hand and left -hand continuity

A function is said to be continuous on the right at x=a if  $\lim_{x\rightarrow a^+}\ f\left(x\right)\ =\ f\left(a\right)$  .

A function is said to be continuous on the  left at x=a if  $\lim_{x\rightarrow a^-}\ f\left(x\right)\ =f\left(a\right)\$  .

eg

Consider f(x)  $\left\{\frac{2x+1\ ,\ x\ge3}{4\ ,\ x<3}\right\}$  

 $\lim_{x\rightarrow3^-}\ f\left(x\right)\ =4$  
 $\lim_{x\rightarrow3^+}\ f\left(x\right)=7$  
 $f\left(3\right)=7$  
f(x) is continuous on the right at x =3 .
f(x) is not continuous on the left at x=3 . Since  $\lim_{x\rightarrow3^-}\ f\left(x\right)\ \ne f\left(3\right)$  

7. Continuity in an interval.

f(x) is said to be continuous in an interval if it is continuous at all points in the interval. In particular, if f(x) is defined on the closed interval [a,b] , then f(x) is continuous if and only if

 $\lim_{x\rightarrow x_0}f\left(x\right)=f\left(x_0\right)\ for\ a\le x_0\le b,$  
 $\lim_{x\rightarrow a^+}=f\left(a\right)\ and\ \lim_{x\rightarrow b^-}\ =f\left(b\right)$  

Note
[ ]  - closed interval using  $\le and\ \ge$  signs
( ) - open interval using  $<\ and\ >signs$  

8. Theorems on continuity

(a) If f(x) and g(x) are both continuous at x = a , then so to are the functions :
f(x) + g(x) , f(x) - g(x) , f(x) * g(x) and f(x) / g(x) , the last being true only if g(a) is not equal to 0.

(b) The following functions are continuous in every interval :
* all polynomials * sin x and cos x * $a^x,\ a>0$  

Example 1

Use the given graph of y = f(x) to state the value of each quantity, if it exists. If it does not exist explain why.
(a)  $\lim_{x\rightarrow2^-}\ f\left(x\right)$  
(b) $\lim_{x\rightarrow2^+}\ f\left(x\right)$  
(c)  $\lim_{x\rightarrow2}\ f\left(x\right)$  
(d)  $f\left(2\right)$  
(e)  $\lim_{x\rightarrow4}\ f\left(x\right)$  
(f)  $f\left(4\right)$  

Answers

(a)  $\lim_{x\rightarrow2^-}f\left(x\right)\ =3$  

(b) $\lim_{x\rightarrow2^+}f\left(x\right)$  
(c)  $\lim_{x\rightarrow2}\ f\left(x\right)\$  - does not exist because the left hand and right hand limits are not equal.
(d) f(2) = 3
(e)  $\lim_{x\rightarrow4}\ f\left(x\right)\ =4$  
 (f) f(4) - does not exist 

Example 2

For the function f whose graph is given, state the value of each quantity, if it exist. If i does not exist state why.
(a) $\lim_{x\rightarrow1}\ f\left(x\right)$  
(b)  $\lim_{x\rightarrow3^-}\ f\left(x\right)$  
(c)  $\lim_{x\rightarrow3^+}\ f\left(x\right)$  
(d) $\lim_{x\rightarrow3}\ f\left(x\right)$  
(e)  $f\left(3\right)$  
 

Answers

(a)  $\lim_{x\rightarrow1}\ f\left(x\right)\ =2$  
(b) $\lim_{x\rightarrow3^-}\ f\left(x\right)\ =1$  
(c)  $\lim_{x\rightarrow3^+}\ f\left(x\right)\ =4$  
(d)  $\lim_{x\rightarrow3}\ f\left(x\right)\$  - No limit. It does not exist because the left and right hand limits are not the same.
(e)  $f\left(3\right)\ =3$  
 

Example 3

The g(x) is defined as  $g\left(x\right)\ =\ \left\{\frac{x^{2\ }-3,\ x\ge4}{2x+k\ x<4}\right\}$  

Find the value of 

(a)  $\lim_{x\rightarrow4^-}\ g\left(x\right)$  
(b)  $\lim_{x\rightarrow4^+}g\left(x\right)$  
(c) f(4) 

Answers

(a)  $\lim_{x\rightarrow4^-}2x+k$                   $\left(d\right)\lim_{x\rightarrow4^-}\ g\left(x\right)\ =\lim_{x\rightarrow4}+g\left(x\right)\ =g\left(4\right)$  
 $=2\left(4\right)+k$                                   $8+k=13=13$  
 $=8+k$                                         $k=13-8$  
                                                               $k=5$                                                                                      
(b )  $\lim_{x\rightarrow4^+}x^2-3$  
 $=\left(4\right)^2-3$  
 $=13$  

(c) g(4) =  $(4)^2-3$  

=13