Mathematics
Posted by Asok K Nadhani on

Limits & Continuity in Calculus- Basic Concepts

Definition of Limits in Calculus

In mathematics, a Limit is the value that a function or sequence “approaches” as the input or index approaches some value. Limits are essential to calculus (and mathematical analysis in general) and are used to define continuity, derivatives, and integrals.

Here, we will dicuss about the Basic Cnecpts of following terms, related to limits and Contnuity.

  • Limit of a Function
  • Evaluation of Algebraic Limits
  • Direct Substitution Method
  • Factorisation Method
  • Rationalisation Method
  • Continuity of a mathematical function
  • Continuous Function

Basic Concept of Limits in Calculus

Concept of Limit may be mathematically expressed as

\displaystyle \underset{{n\to c}}{\mathop{{\lim }}}\,f(c)=L and may be read as “the limit of f (n) equals L, as n approaches c “. Here “lim” indicates limit, and the fact that function f(n) approaches the limit L, as n approaches c is represented by the right arrow (n→c)

Limit of a Function

Let f be function; if f(x) is defined for all real values of x, in a neighborhood of a fixed number a, and f(x) approaches a fixed number l as x → a from either side of a, Then l is called the limit of f (x) as x → a and we write \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,f(x)=I

In this function, as the value of x approaches nearer and nearer to a, then the value of f(x) also comes nearer and nearer to l.

If f(x) approaches a limit l as x → α, then we write \displaystyle \underset{{x\to \alpha }}{\mathop{{\lim }}}\,f(x)=I

In some cases, it may so happen that as x approaches a value, the value of the function f(x) may become larger and larger without any limit.

This is symbolically written as \displaystyle \underset{{x\to I}}{\mathop{{\lim }}}\,f(x)=\alpha

Fundamental Results of Limits of Function

Let f and g be two real functions such that

\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,f(x)=I\text{ and }\underset{{x\to a}}{\mathop{{\lim }}}\,g(x)=m where l and m are finite, then

\displaystyle 1.\text{ }\underset{{x\to a}}{\mathop{{\lim }}}\,\left[ {f(x)+g(x)} \right]=\underset{{x\to a}}{\mathop{{\lim }}}\,f(x)\pm \underset{{x\to a}}{\mathop{{\lim }}}\,g(x)=I+m

\displaystyle 2.\text{ }\underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{f(x)}}{{g(x)}}=\frac{{\underset{{x\to a}}{\mathop{{\lim }}}\,f(x)}}{{\underset{{x\to a}}{\mathop{{\lim }}}\,g(x)}}\text{ provided }\underset{{x\to a}}{\mathop{{\lim }}}\,g(x)=0

Computing Limits of Functions

(i)    For \displaystyle {\underset{{x\to a}}{\mathop{{\lim }}}\,f(x)}, put x = a directly in the expression of f(x) provided we do not get a form like  (or denominator = 0) and hence find the limiting value.

(ii)    In \displaystyle {\underset{{x\to a}}{\mathop{{\lim }}}\,f(x)}, if the expression of f(x) attains the form  (or denominator = 0) after putting x = a, then put x = a + h (h is a small number) so that x → a as h → 0 i.e. \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,f(x)=\underset{{h\to 0}}{\mathop{{\lim }}}\,f(a+h). Now after simplifying the expression put h = 0 and hence find the value.

Evaluation of Algebraic Limits

Algebraic limit may be evaluated on functions that do not involve trigonometric, inverse trigonometric, exponential and logarithmic functions, by 1. Direct Substitution method, 2.Factorisation method, 3.Rationalisation method, 4. Method of evaluating algebraic limits when x approaches infinity, 5. Using some standard results.

Limits of Simple Functions- Problems Solutions and Examples

Here we explain simple ways of solving problems related to Limits

Ex. 1 : Find \displaystyle \underset{{x\to 1}}{\mathop{{Lim}}}\,\frac{{\left( {3x-1} \right)}}{{\left( {2x-1} \right)}}\

Here, for x = 1, (3x – 1) / (2x – 1) does not attain the form 0/0, so we may put x = 1 in the expression (3x – 1) / (2x – 1) and find the value =[{(3 x 1)} / {(2 x 1)-1}]=2/1=2.

Ex. 2 : Find \displaystyle \underset{{x\to 2}}{\mathop{{Lim}}}\,\frac{{\left( {{{x}^{3}}-8} \right)}}{{\left( {x-2} \right)}}\

Here, for x = 2, the expression (x3 – 8) / (x – 2), takes the form 0/0, so we may put x = 2 + h to get

Evaluation of Limits by Direct Substitution Method

Direct substitution method is used when we get a finite number on direct substitution of the point in the given expression.

Evaluate \displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,\left( {7x+11} \right)
\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,\left( {7x+11} \right)=7.2+11=\left( {14+11} \right)=25

Evaluation of Limits by Factorisation Method

Factorisation method is used when the function of the form  f(x) / g(x) takes indeterminate form.

If by substituting x = a the rational function of the form  f(x) / g(x), the function takes the form 0/0 or \displaystyle \infty /\infty then then (x – a ) is a factor of both f(x) and g(x).

In such case we factorise the numerator and denominator and then cancel the common factor (x – a). After canceling the common factor (x – a ) we again put x =a in the given expression and see whether we get a meaningful number or not. This process is repeated till we get a meaningful number

Ex. \displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}-6x+9}}{{x-3}} At x = 3 the function is undefined (division by zero). While taking the limit as
x → 3 the function is defined near the number 3 because when x  3, x  cannot be exactly equal to 3 i.e. x − 3  0 and consequently division by x − 3 is permissible.

\displaystyle \underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}-6x+9}}{{x-3}} \displaystyle =\underset{{x\to 3}}{\mathop{{\lim }}}\,\frac{{{{{\left( {x-3} \right)}}^{2}}}}{{x-3}}\underset{{x\to 3}}{\mathop{{\lim }}}\,\left( {x-3} \right)=3-3=0

Evaluation of Limits by Rationalisation Method

Rationalisation  method  is used when either the numerator or denominator or both involve expression consisting of square roots and upon substituting of the value of x, the rational expression takes indeterminate form

Ex.1: Evaluate \displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{{\sqrt{{x-4}}}}{{x-16}}

Step 1.  \displaystyle \frac{{\sqrt{{x-4}}}}{{x-16}}=\frac{{\sqrt{{x-4}}}}{{\left( {\sqrt{{x+4}}} \right)\left( {\sqrt{{x-4}}} \right)}}=\frac{1}{{\left( {\sqrt{{x+4}}} \right)}}

Step 2. \displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{{\sqrt{{x-4}}}}{{x-16}}=\underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{1}{{\sqrt{{x+4}}}}=\frac{1}{{\sqrt{8}}}

Ex. 2: Evaluate

The expression  takes the form 0/0 when x = 0

 Multiplying numerator & denominator by (which is ≠ 0 when x = 0)

Evaluate : \displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\left[ {\left( {{{x}^{2}}-3x+2} \right)/\left( {{{x}^{2}}-4x+3} \right)} \right]

 [as x2 – 3x + 2 = x2 – 2x – x + 2 = x(x –2) – 1(x – 2) = (x – 2)(x – 1) & x2 – 4x + 3 = x2 – 3x – x + 3 = x(x – 3) – 1(x – 3) = (x – 3) (x – 1)]

\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\left[ {\left( {x-2} \right)/\left( {x-3} \right)} \right]=\left( {1-2} \right)/\left( {1-3} \right)=1/2

If we put x = 1, the function takes the form 0/0 and becomes undefined. Further limit of denominator is zero. Hence we factorise and cancel by factor (x – 1) which ≠ 0, as x → 1, we evaluate the value.

Evaluating algebraic limits when  \displaystyle \mathbf{x}\to \alpha

Following steps are used to evaluate algebraic limit when value of x tends to infinity.

Step 1 : Write down the given expression in the form of a rational function i.e. f(x) / g(x) if it is not in  that form.

Step 2 : If k is the highest power of x in the numerator and denominator both, then divide each term in numerator and denominator by xk

     Step 3 : Use the result \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\left( {1/n} \right)=0 when n > 0

     Evaluate : \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\left[ {\frac{{\left( {5-2{{x}^{2}}} \right)}}{{\left( {3x+5{{x}^{2}}} \right)}}} \right]

(dividing numerator & denominator byx2)

 = \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\left( {5{{u}^{2}}-2} \right)/\left( {3u+5} \right), (substituting u = 1/x, so, when x → ∞ , u → 0) = –2/5

Evaluating algebraic limits in Standard Forms

Following are some standard results involving limits which are frequently used to evaluate limits

I. \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{{{x}^{n}}-{{a}^{n}}}}{{x-a}}=n{{a}^{{n-1}}},\ \ a>0                          II.     \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{a}^{x}}-1}}{x}={{\log }_{e}}a

III.    \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\log \left( {1+x} \right)}}{x}=1

                                 iv.    \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{e}^{x}}-1}}{x}=1

v.     \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{{\left( {1+x} \right)}}^{n}}-1}}{x}=n                                vi.    \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,{{e}^{x}}=1

vii.   \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,{{x}^{n}}=0

Where, –1 < x < 1.

Evaluate \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{e}^{x}}-{{e}^{{-x}}}}}{x}

Step 1 : \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{e}^{x}}\left( {{{e}^{x}}-{{e}^{{-x}}}} \right)}}{{{{e}^{x}},x}}                

(multiplying numerator & denominator by ex

Step 2 : \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{{{e}^{{2x}}}-1}}{x}x\frac{1}{{{{e}^{x}}}}

Let 2x = z,  Since x → 0, z → 0

Step 3 : \displaystyle \underset{{z\to 0}}{\mathop{{\lim }}}\,\frac{{{{e}^{z}}-1}}{{\frac{z}{2}\times \frac{1}{1}}}                             

Step 4 : \displaystyle 2\cdot \underset{{z\to 0}}{\mathop{{\lim }}}\,\frac{{{{e}^{z}}-1}}{z}x\frac{1}{1}

Step 5 : 2 × 1 × 1 = 2\displaystyle \left[ {as\ \ \underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {{{e}^{x}}-1} \right)/x=1\ \ \And \ \ \underset{{x\to 0}}{\mathop{{\lim }}}\,{{e}^{x}}=1} \right]

Continuity of an algebraic function

A function f(x) is said to be continuous at x = a if and only if

(i)    f(x) is defined at x = a

(ii)    \displaystyle \underset{{x\to a-}}{\mathop{{\lim }}}\,f\left( x \right)=\underset{{x\to a+}}{\mathop{{\lim }}}\,f\left( x \right) both left-hand and right-hand limits exist and are equal.

(iii)   \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,f\left( x \right)=f\left( a \right) limiting value of the function must be equal to its function value x = a.

Properties of Continuous Function

1. Sum of continuous functions is continuous. 2. Difference of continuous functions is continuous.

3. Product of continuous functions is continuous. 4. Quotient of continuous functions is continuous at all points x where the Denominator is not zero. 5. Functional Composition of continuous functions is continuous at all points x where the composition is properly defined. 6. Any Polynomial is continuous for all values of x.

Function ex and Trigonometry functions Sin x & Cos x are continuous for all values of x.

Determine whether the function ƒ(x)  = x2  –  4x  + 3 is continuous at x = 2,

(i)    f(x) = (x2 – 4x + 3), f(x) is defined for x = 2,

(ii)    \displaystyle \underset{{x\to a-}}{\mathop{{\lim }}}\,f\left( x \right)=f\left( a \right)=\underset{{x\to a}}{\mathop{{\lim }}}\,\left( {{{x}^{2}}-4x+3} \right)={{2}^{2}}-\left( {4\times 2} \right)+3=-1 (satisfied)

(iii)   \displaystyle \underset{{x\to a+}}{\mathop{{\lim }}}\,f\left( x \right)=f\left( a \right)=\underset{{x\to a}}{\mathop{{\lim }}}\,\left( {{{x}^{2}}-4x+3} \right)=\underset{{h\to 0}}{\mathop{{\lim }}}\,{{\left( {a-h} \right)}^{2}}-4\underset{{h\to 0}}{\mathop{{\lim }}}\,\left( {a-h} \right)+\underset{{h\to 0}}{\mathop{{\lim }}}\,\left( 3 \right)

       = (2 + 0)2 – 4(2 + 0) + 3 = –1, and f(a) = f(2) = –1 calculated as above

All the above three values are definite and identical as well. Hence, this characteristic is also satisfied.

From the above analysis we find that the value of the function at x = 2, left and right hand limits exist, are finite and identical .

Hence, the given function is continuous at the point x = 2.

It may be noted that the above function is continuous in any interval both for positive and negative values of x.

Continuous Function – Problems and Solutions

Verify continuity of algebraic function

Ex. 1: Verify whether f(x) = |x| continuous at x = 0

     |x| = x, when x > 0, |x| = 0 when x = 0, |x| = –x when x < 0

\displaystyle \underset{{x\to 0-}}{\mathop{{\lim }}}\,f\left( x \right)=\underset{{x\to 0-}}{\mathop{{\lim }}}\,f\left( {-x} \right)=0\ \ and\ \ \underset{{x\to 0+}}{\mathop{{\lim }}}\,f\left( x \right)=\underset{{x\to 0+}}{\mathop{{\lim }}}\,x=0

     Hence lim f(x) = 0 = f(0). So f(x) is continuous at x = 0

Ex. 2 : Show that f(x) = (3x2 – x + 2) is continuous at x = 1

(i)    \displaystyle \underset{{x\to 1+}}{\mathop{{\lim }}}\,f\left( x \right)=\underset{{x\to 1+}}{\mathop{{\lim }}}\,\left( {3{{x}^{2}}-x+2} \right)=\underset{{h\to 0}}{\mathop{{\lim }}}\,\left[ {3{{{\left( {1+h} \right)}}^{2}}-\left( {1+h} \right)+2} \right]=\underset{{h\to 0}}{\mathop{{\lim }}}\,\left( {3{{h}^{2}}+5h+4} \right)

       = 3 × 0 + 5 × 0 + 4 = 4

(ii)    \displaystyle \underset{{x\to 1-}}{\mathop{{\lim }}}\,f\left( x \right)=\underset{{x\to 1-}}{\mathop{{\lim }}}\,\left( {3{{x}^{2}}-x+2} \right)=\underset{{h\to 0}}{\mathop{{\lim }}}\,\left[ {3{{{\left( {1-h} \right)}}^{2}}-\left( {1-h} \right)+2} \right]=\underset{{h\to 0}}{\mathop{{\lim }}}\,\left( {3{{h}^{2}}-5h+4} \right)

       = 3 × 0 + –5 × 0 + 4 = 4

Ex. 3 : Find the points of discontinuity of function  f(x) = [(x2 + 2x + 5) / (x2 − 3x + 2)]

On factorization of the  denominator, we get f(x) = [(x2 + 2x + 5) / {(x – 1) (x – 2)}]

For x =1 and x = 2 the denominator becomes zero, and the function f(x) is undefined at x = 1 and x = 2. Hence the points of discontinuity are x = 1 and x = 2.

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