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Example 5  Prove that

Solution :  f(x) = , l = 2 and a = 1.

Given e > 0 we must find d > 0

such that | - 2 | < e , 0 < | x - 1 | < d ---------- (1)

Multiplying the left side by we see

That

or    ------ (2)

If we agree to take d < 1, then | x - 1 | < d Þ x Î ( 0, 2 )

and hence + 2 > Ö3 + 2. Thus, if | x - 1 | < d < 1 then

If we pick d = min ( 1, Î (Ö3 + 2 )) then

hence (2) holds. Thus

Note :"One-sided" limits always exist for an important class of functions - namely, monotone functions

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Note : From our previous discussion and problems, we can consider the following four possible behaviors of functions x ® a.

(1)   f(x) may tend to a finite limit , say l

(2)   f(x) may tend a

(3)   f(x) may tend -a

(4)   f(x) may neither tend to l nor ¹ a

In the above possibilities (2) and (3) comes under "infinite limits." Some functions approaches in the positive or negative direction (i.e. increasing or decreasing without bound) closer to certain values for an independent variable. When this takes place, the function is said to have an infinite limit; you write

Then the function has a vertical 'asymptotes' at x = a. If either of the above limits holds true.

Limit of a function ' f ' as x ® ¥

Here we suggest the following four possible behaviours of a function 'f' as x® ¥.

(1)   f(x) may tend to finite limit l

(2)   f(x) may tend ¥

(3)   f(x) may tend -¥

(4)   f(x) may neither tend to a finite limit l nor ¹ ¥

Index

2.1 Modulus
2.2 Inequalities
2.3 Limits Of Functions
2.4 Left Hand And Right Hand Limits
2.5 Theorems On The Algebra Of Limits
2.6 Evaluating Limits
2.7 Limits Of Trigonometric Functions
2.8 The Exponential Limits

Chapter 3