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# Limits of algebraic functions

Module by: Sunil Kumar Singh. E-mail the author

Algebraic expressions comprise of polynomials, surds and rational functions. For evaluation of limits of algebraic functions, the main strategy is to work expression such that we get a form which is not indeterminate. Generally, it helps to know “indeterminate form” of expression as it is transformed in each step of evaluation process. The moment we get a determinate form, the limit of the algebraic expression is obtained by plugging limiting value of x in the expression. The approach to transform or change expression depends on whether independent variable approaches finite values or infinity.

The point of limit determines the way we approach evaluation of limit of a function. The treatment of limits involving independent variable tending to infinity is different and as such we need to distinguish these limits from others. Thus, there are two categories of limits being evaluated :

1: Limits of algebraic function when variable tends to finite value.

2: Limits of algebraic function when variable tends to infinite

## Limits of algebraic function when variable tends to finite value

In essence, we shall be using following three techniques to determine limit of algebraic expressions when variable is approaching finite value – not infinity. These methods are :

1: Simplification or rationalization (for radical functions)

2: Using standard limit form

3: Canceling linear factors (for rational function)

We should be aware that if given function is in determinate form, then we need not process the expression and obtain limit simply by plugging limiting value of x in the expression. Some problems can be alternatively solved using either of above methods.

### Simplification or rationalization (for radical functions

We simplify or rationalize (if surds are involved) and change indeterminate form to determinate form. We need to check indeterminate forms after each simplification and should stop if expression turns determinate. In addition, we may use following results for rationalizing expressions involving surds :

a b = a b a + b a b = a b a + b a 1 / 3 - b 1 / 3 = a b a 2 / 3 + a 1 / 3 b 1 / 3 + b 2 / 3 a 1 / 3 - b 1 / 3 = a b a 2 / 3 + a 1 / 3 b 1 / 3 + b 2 / 3

#### Example 1

Problem : Determine limit

lim x 1 x - 1 2 x - 1 2 x 2 x 1 lim x 1 x - 1 2 x - 1 2 x 2 x 1

Solution : Here, indeterminate form is 0/0. We simplify to change indeterminate form and find limit,

x - 1 2 x - 1 2 x 2 x 1 = x - 1 2 x - 1 x - 1 2 x + 1 = 2 x - 1 x + 1 2 x + 1 x - 1 2 x - 1 2 x 2 x 1 = x - 1 2 x - 1 x - 1 2 x + 1 = 2 x - 1 x + 1 2 x + 1

This is determinate form. Plugging “1” for x, we have :

L = 1 6 L = 1 6

#### Example 2

Problem : Determine limit

lim x 0 1 x 8 + x 1 3 1 2 x lim x 0 1 x 8 + x 1 3 1 2 x

Solution : The indeterminate form is ∞-∞. Simplifying, we have :

f x = 2 - 8 + x 1 3 2 x 8 + x 1 3 f x = 2 - 8 + x 1 3 2 x 8 + x 1 3

We know that :

a 1 / 3 - b 1 / 3 = a b a 2 / 3 + a 1 / 3 b 1 / 3 + b 2 / 3 a 1 / 3 - b 1 / 3 = a b a 2 / 3 + a 1 / 3 b 1 / 3 + b 2 / 3

Using this identity :

2 - 8 + x 1 3 = 8 1 3 - 8 + x 1 3 2 - 8 + x 1 3 = 8 1 3 - 8 + x 1 3 = 8 8 x 8 2 3 + 8 1 3 8 1 3 + 8 2 3 = 8 8 x 8 2 3 + 8 1 3 8 1 3 + 8 2 3

Substituting in the given expression,

= x 2 x 8 + x 1 3 8 2 3 + 8 1 3 8 1 3 + 8 2 3 = x 2 x 8 + x 1 3 8 2 3 + 8 1 3 8 1 3 + 8 2 3 = 1 2 8 + x 1 3 8 2 3 + 8 1 3 8 1 3 + 8 2 3 = 1 2 8 + x 1 3 8 2 3 + 8 1 3 8 1 3 + 8 2 3

This is a determinate form. Plugging “0” for x,

L = 1 2 X 8 1 3 8 2 3 + 8 1 3 8 1 3 + 8 2 3 = 1 48 L = 1 2 X 8 1 3 8 2 3 + 8 1 3 8 1 3 + 8 2 3 = 1 48

#### Example 3

Problem : Determine limit :

lim x 0 1 x 2 - 1 - x x lim x 0 1 x 2 - 1 - x x

Solution : Here, indeterminate form is 0/0. We simplify to change indeterminate form and find limit,

1 x 2 - 1 - x x = 1 x 2 - 1 + x x { 1 x 2 + 1 - x } = 1 x { 1 x 2 + 1 - x } 1 x 2 - 1 - x x = 1 x 2 - 1 + x x { 1 x 2 + 1 - x } = 1 x { 1 x 2 + 1 - x }

This simplified form is not indeterminate. Plugging “0” for “x” :

L = 1 2 L = 1 2

### Using standard limit form

There is an important algebraic form which is used as standard form. The standard form is (n is rational number) :

lim x 0 x n a n x - a = n a n 1 lim x 0 x n a n x - a = n a n 1

For rational “n”, the expansion of the expression in the limit is given by :

x n a n x - a = x n 1 + a x n 2 + a 2 x n 3 + . . + a n 1 x n a n x - a = x n 1 + a x n 2 + a 2 x n 3 + . . + a n 1

The expression on the right hand side of the equation is not indeterminate. Thus, limit is obtained simply by plugging “a” for “x” :

L = a n 1 + a X a n 2 + a 2 a n 3 + . . + a n 1 L = a n 1 + a X a n 2 + a 2 a n 3 + . . + a n 1 L = n a n 1 L = n a n 1

#### Example 4

Problem : Determine limit

lim x a x 3 / 2 a 3 / 2 x 1 / 2 a 1 / 2 lim x a x 3 / 2 a 3 / 2 x 1 / 2 a 1 / 2

Solution : Here, indeterminate form is 0/0. Substituting y = x 1 / 2 and b = a 1 / 2 y = x 1 / 2 and b = a 1 / 2 .

x 3 / 2 a 3 / 2 x 1 / 2 a 1 / 2 = y 3 b 3 y b x 3 / 2 a 3 / 2 x 1 / 2 a 1 / 2 = y 3 b 3 y b

As x - > a , y - > b x - > a , y - > b . Using formula,

L = 3 b 2 = 3 a 1 / 2 2 = 3 a L = 3 b 2 = 3 a 1 / 2 2 = 3 a

#### Example 5

Problem : Determine limit

lim h 0 x + h 1 / n - x 1 / n h lim h 0 x + h 1 / n - x 1 / n h

Solution : Here, indeterminate form is 0/0. Rearranging and using formulae,

lim h 0 x + h 1 / n - x 1 / n x + h x lim h 0 x + h 1 / n - x 1 / n x + h x

As h 0, x + h x h 0, x + h x . Using formulae,

L = 1 n x 1 n - 1 L = 1 n x 1 n - 1

### Canceling linear factors (for rational function)

If expression is a rational function, then it is likely that both numerator and denominator become zero at x=a such that given expression has 0/0 indeterminate form. Clearly, then (x-a) is a factor of both numerator and denominator. Canceling common linear factor, we get determinate form. We evaluate limit by plugging limiting value of x in the expression.

#### Example 6

Problem : Determine limit :

lim x 3 x 3 3 x 2 x + 3 x 2 x 6 lim x 3 x 3 3 x 2 x + 3 x 2 x 6

Solution : Here, indeterminate form is 0/0. The numerator and the denominator individually tend to 0 as x →3. It means (x-3) is factor of both numerator and denominator. Dividing polynomials (long method or otherwise), we find the quotient. We replace expression in terms of linear factor and quotient :

x 3 3 x 2 x + 3 x 2 x 6 = x 3 x 2 1 x 3 x + 2 x 3 3 x 2 x + 3 x 2 x 6 = x 3 x 2 1 x 3 x + 2 = x 2 1 x + 2 = x 2 1 x + 2

This is determinate form. Plugging “3” for “x”, we have :

L = 8 5 L = 8 5

## Limits of algebraic function when variable tends to infinity

In this case, we are required to estimate behavior of expression when variable approaches very large positive or negative value. The basic idea is to obtain each term in “reciprocal form”. As variable approaches infinity, the reciprocal term approaches zero. There are two methods. However, we need to simplify expression before using either of these methods.

We should also note that these two methods are completely equivalent. We can use either of two methods to evaluate limits. Application of a method is a matter of choice and ease.

1: Dividing each term by highest power of variable

We divide each term of numerator and denominator by x raised to highest positive power in the given expression. Then, we use following limit,

x , c x n 0 ; n > 0 x , c x n 0 ; n > 0

2: Take out highest power of variable

We take out x raised to highest power from numerator and denominator separately. The highest power variable is considered separately for numerator and denominator. This is unlike previous case in which we use highest power variable of the whole expression. Finally, we evaluate resulting expression using limit rules specified above for the reciprocals and using following additional limits,

When x , When x , x n 0 if n < 0 x n 0 if n < 0 x n 1 if n = 0 x n 1 if n = 0 x n if n > 0 x n if n > 0

We should again emphasize that two methods are essentially equivalent.

Dividing radical by variable poses difficulty when variable represents negative value. Such is the case, when we are evaluating limit in which variable is approaching negative infinity. Clearly, variable has negative value in such cases. We use following rule :

If x < 0, then x = - x 2 If x < 0, then x = - x 2

In order to understand working of this rule, let us consider a radical :

x 2 + 4 x x 2 + 4 x

We are required to divide the radical by x, when it is known to be negative. Following the fact stated above,

x 2 + 4 x x = x 2 + 4 x x 2 = 1 + 4 x x 2 + 4 x x = x 2 + 4 x x 2 = 1 + 4 x

### Example 7

Problem : Determine limit :

lim x x 4 + 2 x 3 + 3 2 x 4 x + 2 lim x x 4 + 2 x 3 + 3 2 x 4 x + 2

Solution : Here, indeterminate form is ∞/∞. Dividing each term by x 4 x 4

x 4 + 2 x 3 + 3 2 x 4 x + 2 = 1 + 2 x + 3 x 4 2 1 x 3 + 2 x 4 x 4 + 2 x 3 + 3 2 x 4 x + 2 = 1 + 2 x + 3 x 4 2 1 x 3 + 2 x 4

This is determinate form. As x , 2 / x 0, 3 x 4 0, Numerator 1 x , 2 / x 0, 3 x 4 0, Numerator 1 and as x , 1 x 3 0, 2 x 4 0, Denominator 2 x , 1 x 3 0, 2 x 4 0, Denominator 2 . Hence, limit is :

L = 1 2 L = 1 2

Alternatively, we can employ second method to evaluate limit. Taking out x 4 x 4

x 4 + 2 x 3 + 3 2 x 4 x + 2 = x 4 1 + 2 x + 3 x 4 x 4 2 1 x 3 + 2 x 4 x 4 + 2 x 3 + 3 2 x 4 x + 2 = x 4 1 + 2 x + 3 x 4 x 4 2 1 x 3 + 2 x 4

L = 1 2 L = 1 2

### Example 8

Problem : Determine limit :

lim x a n + b n a n b n ; a > b > 1 lim x a n + b n a n b n ; a > b > 1

Solution : Here, indeterminate form is ∞/∞. By inspection, we see that a/b > 1 and b/a < 1. As we know x→∞, c x c x → 0 if c < 1. Hence, we are required to get terms in the form b/a raised to some power. Dividing numerator and denominator by a n a n , we have :

a n + b n a n b n = 1 + b a n 1 b a n a n + b n a n b n = 1 + b a n 1 b a n

This is determinate form. As n , b / a n 0 n , b / a n 0 . Hence,

L = 1 L = 1

### Example 9

Problem : Determine limit :

lim x 1 2 n 3 + 2 2 n 3 + 3 2 n 3 + + n 2 n 3 lim x 1 2 n 3 + 2 2 n 3 + 3 2 n 3 + + n 2 n 3

Solution : The indeterminate form is ∞/∞. Writing expression of sum of square of natural numbers, we have :

1 2 n 3 + 2 2 n 3 + 3 2 n 3 + + n 2 n 3 = 1 2 + 2 2 + 3 2 + + n 2 n 3 1 2 n 3 + 2 2 n 3 + 3 2 n 3 + + n 2 n 3 = 1 2 + 2 2 + 3 2 + + n 2 n 3 = n n + 1 2 n + 1 2 n 3 = n + 1 2 n + 1 2 n 2 = 2 + 3 n + 1 n 2 2 = n n + 1 2 n + 1 2 n 3 = n + 1 2 n + 1 2 n 2 = 2 + 3 n + 1 n 2 2

As n , 3 / n , 1 / n 2 0 n , 3 / n , 1 / n 2 0

L = 2 2 = 1 L = 2 2 = 1

### Example 10

Problem : Determine limit :

lim x -∞ { x 2 + 4 x x 2 4 x } lim x -∞ { x 2 + 4 x x 2 4 x }

Solution : Here, indeterminate form is ∞-∞. Rationalizing, we have :

x 2 + 4 x x 2 4 x = 8 x { x 2 + 4 x + x 2 4 x } x 2 + 4 x x 2 4 x = 8 x { x 2 + 4 x + x 2 4 x }

Dividing by x. Note x is negative. Hence x = - x 2 x = - x 2

= 8 { 1 + 4 x 1 4 x } = 8 { 1 + 4 x 1 4 x }

As x - , 4 x 0. x - , 4 x 0. .

L = 4 L = 4

## Exercises

### Exercise 1

Determine limit

lim x 1 1 x 2 1 2 x 4 - 1 lim x 1 1 x 2 1 2 x 4 - 1

#### Solution

Here, indeterminate form is ∞ - ∞. We simplify to change the form of expression from determinate ,

1 x 2 1 2 x 4 - 1 = 1 x 2 1 2 x 2 - 1 x 2 + 1 1 x 2 1 2 x 4 - 1 = 1 x 2 1 2 x 2 - 1 x 2 + 1 = x 2 + 1 - 2 x 2 - 1 x 2 + 1 = x 2 - 1 x 2 - 1 x 2 + 1 = x 2 + 1 - 2 x 2 - 1 x 2 + 1 = x 2 - 1 x 2 - 1 x 2 + 1 = 1 x 2 + 1 = 1 x 2 + 1

This form is determinate. Plugging “1” for “x”, we have :

L = 1 2 L = 1 2

### Exercise 2

Determine limit

lim x x { x + c x } lim x x { x + c x }

#### Solution

Here, indeterminate form is ∞-∞. Rationalizing, we have :

x { x + c x } = c x { x + c + x } x { x + c x } = c x { x + c + x }

Dividing by x x ,

= c { 1 + c x + 1 } = c { 1 + c x + 1 }

As x , c / x 0 x , c / x 0

L = c 2 L = c 2

### Exercise 3

Determine limit

lim x { x - x 2 + x } lim x { x - x 2 + x }

#### Solution

Here, indeterminate form is ∞-∞. Rationalizing surd, we have :

{ x - x 2 + x } = { x - x 2 + x } { x + x 2 + x } { x + x 2 + x } { x - x 2 + x } = { x - x 2 + x } { x + x 2 + x } { x + x 2 + x }

= x 2 - x 2 - x { x + x 2 + x } = - x { x + x 2 + x } = x 2 - x 2 - x { x + x 2 + x } = - x { x + x 2 + x }

Dividing each of terms by x, we have :

= - 1 { 1 + 1 + 1 x } = - 1 { 1 + 1 + 1 x }

This is determinate form. As x->∞, 1/x -> 0.

L = 1 1 + 1 = - 1 2 L = 1 1 + 1 = - 1 2

### Exercise 4

Determine limit :

lim x { x + x - x } lim x { x + x - x }

#### Solution

Here, indeterminate form is ∞-∞. Rationalizing surds, we have :

{ x + x - x } = { x + x - x } { x + x + x } { x + x + x } { x + x - x } = { x + x - x } { x + x + x } { x + x + x }

= x + x - x { x + x + x } = x { x + x + x } = x + x - x { x + x + x } = x { x + x + x }

Dividing numerator and denominator by x x ,

= 1 { 1 + 1 x + 1 } = 1 { 1 + 1 x + 1 }

This is determinate form. As x , 1 x 0 x , 1 x 0 ,

L = 1 2 L = 1 2

### Exercise 5

Determine limit

lim x a x 5 / 2 a 5 / 2 x 1 / 2 a 1 / 2 lim x a x 5 / 2 a 5 / 2 x 1 / 2 a 1 / 2

#### Solution

Here, indeterminate form is 0/0. We put y = x 1 / 2 , b = a 1 / 2 ; x a , y b y = x 1 / 2 , b = a 1 / 2 ; x a , y b .

x 5 / 2 a 5 / 2 x 1 / 2 a 1 / 2 = x 5 a 5 x a x 5 / 2 a 5 / 2 x 1 / 2 a 1 / 2 = x 5 a 5 x a

Using formulae :

L = 5 b 4 L = 5 b 4

L = 5 a 1 / 2 4 L = 5 a 1 / 2 4 L = 5 a 2 L = 5 a 2

### Exercise 6

Determine limit

lim x 1 1 x - 1 2 x 2 - 1 lim x 1 1 x - 1 2 x 2 - 1

#### Solution

Here, indeterminate form is ∞ - ∞. We simplify to change indeterminate form and find limit,

1 x - 1 2 x 2 - 1 = 1 x - 1 2 x - 1 x + 1 1 x - 1 2 x 2 - 1 = 1 x - 1 2 x - 1 x + 1 = x + 1 - 2 x - 1 x + 1 = x - 1 x - 1 x + 1 = x + 1 - 2 x - 1 x + 1 = x - 1 x - 1 x + 1 = 1 x + 1 = 1 x + 1

It is determinate form. Plugging “1” for “x”, we have :

L = 1 2 L = 1 2

### Exercise 7

Determine limit :

lim x x p + x p - 1 + 1 x q + x q - 1 + 1 ; p > 0, q > 0 lim x x p + x p - 1 + 1 x q + x q - 1 + 1 ; p > 0, q > 0

#### Solution

Here, indeterminate form is ∞/∞. We divide each term by x p x p .

x p + x p - 1 + 1 x q + x q - 1 + 1 = x p 1 + 1 x + 1 x p x q 1 + 1 x + 1 x q x p + x p - 1 + 1 x q + x q - 1 + 1 = x p 1 + 1 x + 1 x p x q 1 + 1 x + 1 x q

As x , 1 x and 1 x p 0 x , 1 x and 1 x p 0 , and

x p - q ; if p > q x p - q ; if p > q x p - q 1 ; if p = q x p - q 1 ; if p = q x p - q 0 ; if p < q x p - q 0 ; if p < q

Hence,

L = ; if p > q L = ; if p > q L = 1 ; if p = q L = 1 ; if p = q L = 0 ; if p < q L = 0 ; if p < q

### Exercise 8

Determine limit :

lim x 0 1 + x 3 - 1 3 x + 2 x 2 lim x 0 1 + x 3 - 1 3 x + 2 x 2

#### Solution

Here, indeterminate form is 0/0. Using standard form,

1 + x 3 - 1 3 x + 2 x 2 = 1 + x 3 - 1 1 + x - 1 X x 3 x + 2 x 2 1 + x 3 - 1 3 x + 2 x 2 = 1 + x 3 - 1 1 + x - 1 X x 3 x + 2 x 2 = 1 + x 3 - 1 1 + x - 1 X 1 3 + 2 x = 1 + x 3 - 1 1 + x - 1 X 1 3 + 2 x

This is determinate form. x 0, 1 + x 1 x 0, 1 + x 1

L = 3 X 1 2 3 = 1 L = 3 X 1 2 3 = 1

### Exercise 9

Determine limit :

lim x 2 x - 2 + x - 2 x 2 4 lim x 2 x - 2 + x - 2 x 2 4

#### Solution

Here, indeterminate form is 0/0. We simplify to change indeterminate form and find limit,

x - 2 + x - 2 x 2 4 = x - 2 x 2 4 + x - 2 x 2 4 x - 2 + x - 2 x 2 4 = x - 2 x 2 4 + x - 2 x 2 4 = 1 x + 2 + x - 2 x + 2 x + 2 x 2 4 = 1 x + 2 + x - 2 x + 2 x + 2 x 2 4 = 1 x + 2 + x - 2 x + 2 x + 2 = 1 x + 2 + x - 2 x + 2 x + 2

This is determinate form. Plugging “2” for x, we have :

L = 1 2 L = 1 2

### Exercise 10

Determine limit :

lim x 1 x - 1 x 2 - 1 + x - 1 lim x 1 x - 1 x 2 - 1 + x - 1

#### Solution

Here, indeterminate form is 0/0. Rationalizing surds,

x - 1 x 2 - 1 + x - 1 = x - 1 x - 1 X x - 1 { x 2 - 1 x - 1 X x - 1 } + { x - 1 x - 1 X x - 1 } x - 1 x 2 - 1 + x - 1 = x - 1 x - 1 X x - 1 { x 2 - 1 x - 1 X x - 1 } + { x - 1 x - 1 X x - 1 }

Each term limits to n a n 1 n a n 1 .

This is determinate form.

L = 0 2 + 1 = 0 L = 0 2 + 1 = 0

### Exercise 11

Determine limit

lim x 64 x 1 / 6 2 x 1 / 3 4 lim x 64 x 1 / 6 2 x 1 / 3 4

#### Solution

Here, indeterminate form is 0/0. Using standard formulae, we have :

x 1 / 6 2 x 1 / 3 4 = x 1 / 6 64 1 / 6 x 1 / 3 64 1 / 3 x 1 / 6 2 x 1 / 3 4 = x 1 / 6 64 1 / 6 x 1 / 3 64 1 / 3 = x 1 / 6 64 1 / 6 x 64 x 1 / 3 64 1 / 3 x 64 = x 1 / 6 64 1 / 6 x 64 x 1 / 3 64 1 / 3 x 64

It is determinate form. Evaluating, we have :

L = 1 6 X 64 1 / 6 1 1 3 X 64 1 / 3 1 L = 1 6 X 64 1 / 6 1 1 3 X 64 1 / 3 1

L = 1 4 L = 1 4

### Exercise 12

Determine limit :

lim x 1 { x 2 + 8 - 10 - x 2 } { x 2 + 3 - 5 - x 2 } lim x 1 { x 2 + 8 - 10 - x 2 } { x 2 + 3 - 5 - x 2 }

#### Solution

Here, indeterminate form is ∞-∞. Rationalizing surds,

{ x 2 + 8 - 10 - x 2 } { x 2 + 3 - 5 - x 2 } = { x 2 + 8 - 10 - x 2 } { x 2 + 8 + 10 - x 2 } { x 2 + 3 + 5 - x 2 } { x 2 + 8 + 10 - x 2 } { x 2 + 3 - 5 - x 2 } { x 2 + 3 + 5 - x 2 } { x 2 + 8 - 10 - x 2 } { x 2 + 3 - 5 - x 2 } = { x 2 + 8 - 10 - x 2 } { x 2 + 8 + 10 - x 2 } { x 2 + 3 + 5 - x 2 } { x 2 + 8 + 10 - x 2 } { x 2 + 3 - 5 - x 2 } { x 2 + 3 + 5 - x 2 } = x 2 + 8 10 + x 2 { x 2 + 3 + 5 - x 2 } x 2 + 3 5 + x 2 { x 2 + 8 + 10 - x 2 } = x 2 + 8 10 + x 2 { x 2 + 3 + 5 - x 2 } x 2 + 3 5 + x 2 { x 2 + 8 + 10 - x 2 } = 2 x 2 2 { x 2 + 3 + 5 - x 2 } 2 x 2 2 { x 2 + 8 + 10 - x 2 } = 2 x 2 2 { x 2 + 3 + 5 - x 2 } 2 x 2 2 { x 2 + 8 + 10 - x 2 } = { x 2 + 3 + 5 - x 2 } { x 2 + 8 + 10 - x 2 } = { x 2 + 3 + 5 - x 2 } { x 2 + 8 + 10 - x 2 }

This is in determinate form. Plugging “1” for “x”, we have :

L = 2 + 2 3 + 3 = 2 3 L = 2 + 2 3 + 3 = 2 3

### Exercise 13

Determine limit :

lim x 1 x 7 2 x 5 + 1 x 3 3 x 2 + 2 lim x 1 x 7 2 x 5 + 1 x 3 3 x 2 + 2

#### Solution

Here, indeterminate form is 0/0. The numerator and denominator tend to 0 as x->1. It means (x-1) is factor of both numerator and denominator. Dividing polynomials (long method or otherwise) and using quotient :

x 7 2 x 5 + 1 x 3 3 x 2 + 2 = x - 1 x 6 + x 5 - x 4 - x 3 - x 2 - x - 1 x - 1 x 2 - 2 x - 2 x 7 2 x 5 + 1 x 3 3 x 2 + 2 = x - 1 x 6 + x 5 - x 4 - x 3 - x 2 - x - 1 x - 1 x 2 - 2 x - 2 = x 6 + x 5 - x 4 - x 3 - x 2 - x - 1 x 2 - 2 x - 2 = x 6 + x 5 - x 4 - x 3 - x 2 - x - 1 x 2 - 2 x - 2

This is determinate form. Plugging “1” by “x”, we know :

L = - 3 - 3 = 1 L = - 3 - 3 = 1

### Exercise 14

Determine limit :

lim x a a + 2 x - 3 x 3 a + x - 2 x ; x a lim x a a + 2 x - 3 x 3 a + x - 2 x ; x a

#### Solution

Indeterminate form is 0/0. We simplify the expression to change indeterminate form and find limit,

a + 2 x - 3 x 3 a + x - 2 x = { a + 2 x - 3 x } { a + 2 x + 3 x } { 3 a + x + 2 x } { a + 2 x + 3 x } { 3 a + x - 2 x } { 3 a + x + 2 x } a + 2 x - 3 x 3 a + x - 2 x = { a + 2 x - 3 x } { a + 2 x + 3 x } { 3 a + x + 2 x } { a + 2 x + 3 x } { 3 a + x - 2 x } { 3 a + x + 2 x } = a + 2 x 3 x { 3 a + x + 2 x } 3 a + x - 4 x { a + 2 x + 3 x } = a + 2 x 3 x { 3 a + x + 2 x } 3 a + x - 4 x { a + 2 x + 3 x } = a x { 3 a + x + 2 x } 3 a - x { a + 2 x + 3 x } = a x { 3 a + x + 2 x } 3 a - x { a + 2 x + 3 x } = { 3 a + x + 2 x } 3 { a + 2 x + 3 x } = { 3 a + x + 2 x } 3 { a + 2 x + 3 x }

This is not in indeterminate form. Plugging “a” for “x”

L = { 2 a + 2 a } 3 { 3 a + 3 a } = { 4 a } 6 3 a L = { 2 a + 2 a } 3 { 3 a + 3 a } = { 4 a } 6 3 a L = 2 3 3 L = 2 3 3

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