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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

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 = n a n 1

Problem : Determine limit

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 .

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

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

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

Problem : Determine limit

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

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

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.

Problem : Determine limit :

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 2 1 x + 2

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

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

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

We should again emphasize that two methods are essentially equivalent.

Radical and negative variable

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 :

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Source:  OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64
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