Informative line

Algebraic Rules

Sum Rule

  • The limit of the sum is the sum of limits of two functions.
  • This assumes that \(\lim\limits_{x\rightarrow a}f(x)\) and \(\lim\limits_{x\rightarrow a}g(x)\) exists.

Illustration Questions

Consider the graphs of 'f' and 'g', the value of \(\lim\limits_{x\rightarrow -2}\;\Big(f(x)+g(x)\Big)\)from the graph is

A 10

B –13

C –1

D 11

×

\(\lim\limits_{x\rightarrow -2}\;\Big(f(x)+g(x)\Big)\)

\(= \Big(\lim\limits_{x\rightarrow -2}f(x)\Big)+\Big(\lim\limits_{x\rightarrow -2}g(x)\Big)\) \(\rightarrow\) by sum rule

= 1 + (–2) = –1 \(\rightarrow\) look at the height of both graphs at x = –2

Consider the graphs of 'f' and 'g', the value of \(\lim\limits_{x\rightarrow -2}\;\Big(f(x)+g(x)\Big)\)from the graph is

image
A

10

.

B

–13

C

–1

D

11

Option C is Correct

Difference Rule

  • The limits of the difference is the difference of limits of two functions. 
  • This assumes that \(\lim\limits_{x\rightarrow a}f(x)\) and \(\lim\limits_{x\rightarrow a}g(x)\) exists.

Illustration Questions

Consider the graph of 'f' and 'g', the value of \(\lim\limits_{x\rightarrow 3}\;\Big(f(x)-g(x)\Big)\)from the graph is

A –6

B 10

C 17

D 8

×

\(\lim\limits_{x\rightarrow 3}\;\Big(f(x)-g(x)\Big)\)

\(= \Big(\lim\limits_{x\rightarrow 3}f(x)\Big)-\Big(\lim\limits_{x\rightarrow 3}g(x)\Big)\) \(\rightarrow\) By difference rule

= –2 – 4 = –6 \(\rightarrow\) Look at the height of both graph of 'f' and 'g' at x = 3

Consider the graph of 'f' and 'g', the value of \(\lim\limits_{x\rightarrow 3}\;\Big(f(x)-g(x)\Big)\)from the graph is

image
A

–6

.

B

10

C

17

D

8

Option A is Correct

Product Rule

  • The limit of product equals to the product of limit for two function.

Note: \(\lim\limits_{x\rightarrow a}\;f(x)\) and \(\lim\limits_{x\rightarrow a}\;g(x)\) must exist for the rule to follow.

Illustration Questions

Consider the graph of 'f' and 'g', the value of \(\lim\limits_{x\rightarrow -3}\;\Big(f(x)\;g(x)\Big)\) will be

A –10

B –18

C –6

D 12

×

\(\lim\limits_{x\rightarrow -3}\;\Big(f(x)\;g(x)\Big)\)

\(= \Big(\lim\limits_{x\rightarrow -3}f(x)\Big)×\Big(\lim\limits_{x\rightarrow -3}g(x)\Big)\) \(\rightarrow\) By product rule

= 3 × (– 2) = – 6 \(\rightarrow\) Look at the height of both graph of 'f' and 'g' at x = –3

Consider the graph of 'f' and 'g', the value of \(\lim\limits_{x\rightarrow -3}\;\Big(f(x)\;g(x)\Big)\) will be

image
A

–10

.

B

–18

C

–6

D

12

Option C is Correct

Division Rule

The limit of quotient is the quotient of limits of two functions.

Provided  \(\lim\limits_{x\rightarrow a}\;g(x)\neq0 \)

Note: \(\lim\limits_{x\rightarrow a}\;f(x) \) and \(\lim\limits_{x\rightarrow a}\;g(x) \) must exist for the rule.

Illustration Questions

Consider the graph fo two function 'f' and 'g'. The value of \(\lim\limits_{x\rightarrow -3} \left ( \dfrac {f(x)}{g(x)} \right) \) will be

A – 20

B –8

C 12

D –\(\dfrac {1}{4}\)

×

\(\lim\limits_{x\rightarrow -3} \left ( \dfrac {f(x)}{g(x)} \right) \)

\(=\dfrac {\Big(\lim\limits_{x\rightarrow -3}f(x)\Big)} {\Big(\lim\limits_{x\rightarrow -3} g(x)\Big)}\rightarrow \) By quotient rule or division rule

\(\dfrac {1}{-4}=\dfrac {-1}{4}\Rightarrow\) Look at height of the graph of 'f' and 'g' at x = –3

Consider the graph fo two function 'f' and 'g'. The value of \(\lim\limits_{x\rightarrow -3} \left ( \dfrac {f(x)}{g(x)} \right) \) will be

image
A

– 20

.

B

–8

C

12

D

\(\dfrac {1}{4}\)

Option D is Correct

Other Rules For Finding Limits

  1. \(\lim\limits_{x\rightarrow a}\;(f(x))^n = \lim\limits_{x\rightarrow a}\;(f(x))^n\), where n is a positive integer.
  2. \(\lim\limits_{x\rightarrow a}\;x^n=a^n\), where n is a positive integer.
  3. \(\lim\limits_{x\rightarrow a}\;\sqrt [n] {x}=\sqrt [n] {a}\), where n is a positive integer.
  4. \(\lim\limits_{x\rightarrow a}\;\sqrt [n] {f(x)}=\sqrt [n] {\lim\limits_{a \rightarrow a} f(x)}\) where n is a positive integer.

Illustration Questions

Evaluate \(\lim\limits_{x\rightarrow -3} \left ( \dfrac {2x^2+7x-9}{3x+8} \right)\)

A 12

B –18

C 3/7

D 10

×

\(\lim\limits_{x\rightarrow -3} \left ( \dfrac {2x^2+7x-9}{3x+8} \right)\)=\(\dfrac {\lim\limits_{x\rightarrow -3} \left ( {2x^2+7x-9}\right)} {\lim\limits_{x\rightarrow -3} \left ( {3x+8}\right)}\rightarrow\) By quotient Rule

\(=\dfrac {\left (\lim\limits_{x\rightarrow-3}\;2x^2 \right)+ \left (\lim\limits_{x\rightarrow-3}\;7x \right)- \left (\lim\limits_{x\rightarrow -3}\;9 \right) } {\left (\lim\limits_{x\rightarrow-3}\;3x \right)+ \left (\lim\limits_{x\rightarrow-3}\;8 \right)}\rightarrow\) By sum and difference rules

\(=\dfrac {\left (2\lim\limits_{x\rightarrow-3}\;x^2 \right)+ \left (7\lim\limits_{x\rightarrow-3}\;x \right)- \left (\lim\limits_{x\rightarrow -3}\;9 \right) } {\left (3\lim\limits_{x\rightarrow-3}\;x \right)+ \left (\lim\limits_{x\rightarrow-3}\;8 \right)}\rightarrow\) By product rule

\(\dfrac {2×(-3)^2+7×(-3)-(+9)}{3×(-3)+8}\)

\(=\dfrac {18-21-9}{-9+8}=\dfrac {-12}{-1}=12\) (Use rule 'b')

Evaluate \(\lim\limits_{x\rightarrow -3} \left ( \dfrac {2x^2+7x-9}{3x+8} \right)\)

A

12

.

B

–18

C

3/7

D

10

Option A is Correct

 Rule 'D' 

(a)   \(\lim\limits_{x\rightarrow a}\;(f(x))^n = \lim\limits_{x\rightarrow a}\;(f(x))^n\), where n is a positive integer.

(b)   \(\lim\limits_{x\rightarrow a}\;x^n=a^n\), where n is a positive integer.

(c)   \(\lim\limits_{x\rightarrow a}\;\sqrt [n] {x}=\sqrt [n] {a}\), where n is a positive integer.

(d)   \(\lim\limits_{x\rightarrow a}\;\sqrt [n] {f(x)}=\sqrt [n] {\lim\limits_{x \rightarrow a} f(x)}\) , where n is a positive integer.

Illustration Questions

The value of  \(\lim\limits _ {x\to4} \;\sqrt[3] {2x + 19}\)  is 

A 17

B 3

C 25

D – 2

×

\(\lim\limits _{x\to 4} ( 2x + 19)^{1/3}= \left(\lim\limits_{x\to 4}(2x + 19)\right)^{1/3}\)

\( = \underbrace {(2 × 4 +19 )^{1/3}}_\text{Direct substitution rule } =(8+19)^{1/3} = 27 ^{1/3} = 3\)

The value of  \(\lim\limits _ {x\to4} \;\sqrt[3] {2x + 19}\)  is 

A

17

.

B

3

C

25

D

– 2

Option B is Correct

Practice Now