Informative line

Rules For Different Functions

Learn rules of different functions of direct substitution, limits of absolute value function, rationalize the numerator, Evaluating left and right hand limits and greatest integer function.

Direct Substitution Rule for Limits

  • If 'g' is a polynomial or a rational function, then \(\lim\limits_{x\rightarrow a}g(x)=g(a)\)(directly put the value of 'a' in the function).
  • This assumes that x = a is in the domain of 'g'.

Illustration Questions

Evaluate \(\lim\limits_{x\rightarrow -4} \left( \dfrac {2x^2+3}{4x+7} \right )\)

A \(\dfrac {23}{8}\)

B 17

C \(\dfrac {-35}{9}\)

D \(\dfrac {-23}{11}\)

×

Use the direct substitution rule:

\(\Rightarrow\lim\limits_{x\rightarrow -4} \left( \dfrac {2x^2+3}{4x+7} \right )\)

\(=\dfrac {2×(-4)^2+3}{4×(-4)+7}\)

\(=\dfrac {35}{-9}=\dfrac {-35}{9}\)

Evaluate \(\lim\limits_{x\rightarrow -4} \left( \dfrac {2x^2+3}{4x+7} \right )\)

A

\(\dfrac {23}{8}\)

.

B

17

C

\(\dfrac {-35}{9}\)

D

\(\dfrac {-23}{11}\)

Option C is Correct

Evaluation of Limits which are not Possible by Direct Substitution

Consider,

\(\lim\limits_{x\rightarrow 2}\;\dfrac {x-2}{x^2-5x+6}\) 

  • If we put x = 2, we get \(\dfrac {0}{0}\) which is meaningless, so x = 2 is not in the domain of the function.
  • We cancel the factor of x –2 and then put x = 2. So,

\(\Rightarrow\lim\limits_{x\rightarrow 2}\;\dfrac {x-2}{(x-2)(x-3)} \)

\(= \lim\limits_{x\rightarrow 2}\;\dfrac {1}{(x-3)}\)

\(=\dfrac {1}{2-3}=-1\)

  • Remove \(\dfrac {0}{0}\) form and then use direct substitution rule.

Illustration Questions

The value of  \(\lim\limits_{x\rightarrow 3}\;\dfrac {2x^2+x-21}{x^2-9}\)

A 14

B 2/3

C –18

D 13/6

×

If we put x = 3, we get \(\dfrac {0}{0}\) form, so factorize the numerator and denominator.

\(\Rightarrow\lim\limits_{x\rightarrow 3}\;\dfrac {(x-3)(2x+7)}{(x-3)(x+3)}\)

\(=\lim\limits_{x\rightarrow 3}\;\dfrac {2x+7}{x+3}\)

\(=\dfrac {2×3+7}{3+3}=\dfrac {13}{6}\)

The value of  \(\lim\limits_{x\rightarrow 3}\;\dfrac {2x^2+x-21}{x^2-9}\)

A

14

.

B

2/3

C

–18

D

13/6

Option D is Correct

Limits Involving Rationalization

Consider the limit,

\(\lim\limits_{x\rightarrow 0} \left ( \dfrac {\sqrt {1+x}-1}{x} \right)\)

If we put x = 0, we get \(\dfrac {0}{0}\) form, we rationalize the numerator.

\(\lim\limits_{x\rightarrow 0} \dfrac {\sqrt {1+x}-1}{x} \)\(× \dfrac {\sqrt {1+x}+1} {\sqrt {1+x}+1}\)        (Divide & multiply by \({\sqrt {1+x}+1}\))

\(=\lim\limits_{x\rightarrow 0} \dfrac {{1+x}-1}{x(\sqrt {1+x}+1)} \)

\(=\lim\limits_{x\rightarrow 0} \dfrac {1}{(\sqrt {1+x}+1)}=\dfrac {1}{2} \)

  • For rationalizing \(\sqrt a-\sqrt b\), we use the factor \(\sqrt a+\sqrt b\).

 

Illustration Questions

Evaluate \(\lim\limits_{x\rightarrow 2} \left ( \dfrac {\sqrt {7+x}-3}{x-2} \right)\)

A \(\dfrac {1}{6}\)

B \(\dfrac {2}{3}\)

C \(\dfrac {7}{12}\)

D \(\dfrac {8}{13}\)

×

If we put x = 2 we get \(\dfrac {0}{0}\) form, now rationalize the Numerator and then cancel the factor

\(\lim\limits_{x\rightarrow 2} \left ( \dfrac {\sqrt {7+x}-3}{x-2} \right)\)

=\(\lim\limits_{x\rightarrow 2} \dfrac {(\sqrt {7+x}-3)\;(\sqrt {7+x}+3)}{(x-2)(\sqrt {7+x}+3)}\)

\(=\lim\limits_{x\rightarrow 2} \dfrac {(7+x)-9}{(x-2) \sqrt {7+x}+3)}\)

\(=\lim\limits_{x\rightarrow 2} \dfrac {x-2}{(x-2) (\sqrt {7+x}+3)}\)

\(=\lim\limits_{x\rightarrow 2} \dfrac {1}{ \sqrt {7+x}+3}\)

\(= \dfrac {1}{3+3}=\dfrac {1}{6}\)

As soon as we rationalize we will be able to cancel the common factor which will remove \(\dfrac {0}{0}\) form and allow substitution.

Evaluate \(\lim\limits_{x\rightarrow 2} \left ( \dfrac {\sqrt {7+x}-3}{x-2} \right)\)

A

\(\dfrac {1}{6}\)

.

B

\(\dfrac {2}{3}\)

C

\(\dfrac {7}{12}\)

D

\(\dfrac {8}{13}\)

Option A is Correct

Existence of Limits

We know that, 

\(\lim\limits_{x\rightarrow a}\;f(x)\) exists when,

 \(\lim\limits_{x\rightarrow a^+}\;f(x) = \lim\limits_{x\rightarrow a^-}\;f(x)\)  

\(\Rightarrow\) R.H.L. = L.H.L. = limit of the function.

  • In some functions it is necessary to verify the above fact.

Illustration Questions

Let \(f(x)=\begin{cases} 2-3x &if&x\leq2&...(1)\\ x-6 &if&x>2&...(2) \end{cases}\)      Find the \(\lim\limits_{x\rightarrow 2}\;f(x)\) if it exists.

A –10

B –4

C 6

D 3

×

For \(\lim\limits_{x\rightarrow 2}\;f(x)\) to exist \(\rightarrow \) R.H.L. = L.H.L

Now, R.H.L. = \(\lim\limits_{x\rightarrow 2^+}\;f(x) = \lim\limits_{x\rightarrow 2^+}\;(x-6) \)  (2+ > 2 so rule 1)

= 2 – 6 (Direct substitution)

= -4 

L.H.L. = \(\lim\limits_{x\rightarrow 2^-}\;f(x) = \lim\limits_{x\rightarrow 2^-}\;(2-3x) \)  (2 < 2 so rule 2)

= 2 – 3 × 2 (Direct substitution)

 = 2 – 6

= –4

\(\therefore\)  R.H.L. = L.H.L. = –4 

\(\rightarrow \) Limit exists and equals –4.

Whenever there is a piecewise defined function as above there can be a problem of existence of limit at the point from where definition changes.

Let \(f(x)=\begin{cases} 2-3x &if&x\leq2&...(1)\\ x-6 &if&x>2&...(2) \end{cases}\)      Find the \(\lim\limits_{x\rightarrow 2}\;f(x)\) if it exists.

A

–10

.

B

–4

C

6

D

3

Option B is Correct

Limits Based on Modulus Function

  • \(f(x)=|x|\)

                  \(=x\) if \(x\geq0\)  

                  \(=-x\) if \(x<0\)

  • If the quantity inside the modulus function is positive then we open the modulus. 
  • If the quantity inside the modulus function is negative then we open the modulus with negative sign.

Illustration Questions

Let  \(f(x) = \dfrac {x^2 –5x + 6}{| x – 3 | }\) Evaluate \(\lim\limits_{x\rightarrow 3}f(x)\) if it exists

A –4

B 10

C 16

D Limit does not exist

×

For \(\lim\limits_{x\rightarrow 3}\;f(x)\) exist if, R.H.L. = L.H.L

Now, R.H.L.  \(\Rightarrow\lim\limits_{x\rightarrow 3^+}\;f(x) = \lim\limits_{x\rightarrow 3^+}\;\dfrac {x^2-5x+6}{|x-3|}\)

 \(= \lim\limits_{x\rightarrow 3^+}\;\dfrac {x^2-5x+6}{x-3}\)  (\(3^+ > 3\) \(\Rightarrow\;|x-3|=x-3\))

\(= \lim\limits_{x\rightarrow 3^+}\;\dfrac {(x-2)(x-3)}{x-3}\)

\(= \lim\limits_{x\rightarrow 3^+}\; (x-2)\)

= 3 –2(Direct substitution property)

 = 1 

L.H.L. = \(\lim\limits_{x\rightarrow 3^-}\;f(x) = \lim\limits_{x\rightarrow 3^-}\; \dfrac {x^2-5x+6}{|x-3|}\) 

\(= \lim\limits_{x\rightarrow 3^-}\; \dfrac {x^2-5x+6}{-(x-3)}\)   (3 < 3 \(\Rightarrow\;|x-3|=-(x-3)\))

\(= \lim\limits_{x\rightarrow 3^-}\; \dfrac {(x-2)(x-3)}{-(x-3)}\)

\(= \lim\limits_{x\rightarrow 3^-}\; -(x-2)\)

= –(3 – 2)  (Direct substitution)

= –1

\(\therefore\)  R.H.L. \(\neq\) L.H.L.

\(\rightarrow \) Limit does not exist.

Let  \(f(x) = \dfrac {x^2 –5x + 6}{| x – 3 | }\) Evaluate \(\lim\limits_{x\rightarrow 3}f(x)\) if it exists

A

–4

.

B

10

C

16

D

Limit does not exist

Option D is Correct

Greatest Integer Function

We define, 

\(f(x)=[[x]]\) as greatest integer function, it is the value of greatest integer not greater than x.

For example 

[[5.6]] = 5, [[8]] = 8, [[–3.2 ]] = –4,  \([[\sqrt 3]]=1\)

The graph of this function is as shown below. The greatest integer function of any integer is the integer itself.

Due to the nature of the graph of this function, it is called step function. Observe that the graph is broken at all integral values of x.

Illustration Questions

Let f(x) = [[ 2x ]] + 3, then the value of \(f \left ( \dfrac {5}{3} \right)\) is

A 6

B –8

C 7

D 19

×

\(f\left (\dfrac {5}{3}\right) = \Bigg[\Bigg[ 2×\dfrac {5}{3}\Bigg]\Bigg]+3\)

\(f\left (\dfrac {5}{3}\right) = \Bigg[\Bigg[ \dfrac {10}{3}\Bigg]\Bigg]+3\)

\(f\left (\dfrac {5}{3}\right) = [[3.33]]+3\)

\(\Rightarrow f\left (\dfrac {5}{3}\right) \)

\(=3 + 3\)

\(= 6\)

Let f(x) = [[ 2x ]] + 3, then the value of \(f \left ( \dfrac {5}{3} \right)\) is

A

6

.

B

–8

C

7

D

19

Option A is Correct

Non-Existing Limits of Greatest Integer Function

The Greatest Integer Function does not have a Limit at all Integers

This happens because the values of one side limits are different at all integers.

e.g.

\(\lim\limits_{x\rightarrow 1^+}\;[[x]]=1\), whereas \(\lim\limits_{x\rightarrow 1^-}\;[[x]]=0\)

(1+ = 1.0001)           (1 = 0.9999)

Illustration Questions

Find the value of \(\lim\limits_{x\rightarrow \sqrt {2}}\;[[x^2]]\) if it exists.

A 5

B 7

C –6

D Limit doesn't exist

×

R.H.L. = \(\Rightarrow \lim\limits_{x\rightarrow \sqrt {2}^+}\;[[x^2]]\;\)

\(x\rightarrow \sqrt {2}^+ \)

\(\Rightarrow x^2\rightarrow 2^+\)

\(\therefore\)  \(\lim\limits_{x\rightarrow \sqrt {2}^+}\;[[2^+]]\;=2\)

R.H.L. = \(\Rightarrow \lim\limits_{x\rightarrow \sqrt {2}^-}\;[[x^2]]\;\)

\(x\rightarrow \sqrt {2}^- \)

\(\Rightarrow x^2\rightarrow 2^-\)

\(\therefore\)  \(\lim\limits_{x\rightarrow \sqrt {2}^-}\;[[2^-]]\;=1\)

R.H.L.\(\neq\) L.H.L

\(\rightarrow\) Limit does not exists.

Note that \(\lim\limits_{x\rightarrow a}\;[[g(x)]]\) will not exist at those values of x for which g(x) is an integer.

Find the value of \(\lim\limits_{x\rightarrow \sqrt {2}}\;[[x^2]]\) if it exists.

A

5

.

B

7

C

–6

D

Limit doesn't exist

Option D is Correct

Factorization of Quadratic Expressions

To factorize expression of the form \(ax^2 + bx + c\) ,we think of two numbers whose product is ac and sum is b. Then write b as sum of those two numbers. 

e.g.

\(2x^2 + 9x -11\)    ( sum = 9, product = – 22  \(\therefore\) numbers are 11,–2 )

\( = 2x^2 +11x - 2x - 11\)

\( = x (2x +11) - 1(2x +11)\)

\(=(x-1)(2x + 11) \)

Illustration Questions

Factorize \(f(x) = 2x^2 + 7x - 9\) 

A \((x-1)(2x+9)\)

B \((3x + 8)(x+1)\)

C \((3x + 1)(x+ 4)\)

D \((3x + 5)(2x-1)\)

×

\(f(x) = 2x^2 + 7 x-9 \)

Think of two numbers whose product = 2 × – 9 = – 18 ( Coefficient of \(x^2\) × constant )

and whose sum is 7 (coefficient of \(x\) ).

These numbers are – 2 and 9

\(\therefore\; 2x^2 + 7x - 9 = 2x^2 + 9x - 2x - 9\)

\(= x(2x + 9) - 1 (2x + 9)\)

\( = (x-1) (2x + 9)\)

Factorize \(f(x) = 2x^2 + 7x - 9\) 

A

\((x-1)(2x+9)\)

.

B

\((3x + 8)(x+1)\)

C

\((3x + 1)(x+ 4)\)

D

\((3x + 5)(2x-1)\)

Option A is Correct

Practice Now