Tiêu đề LIMIT, CONTINUITY & DIFFERENTIABILITY (Q).pdf ✅

\\ Chapter – 1 The concept of limit is used to discuss the behaviour of a function close to a certain point. e.g., 1 1 ( ) 2    x x f x Clearly the function is not defined at x = 1, but for values close to x = 1 the function can be written as f(x) = x + 1 As x approaches 1 (written as x  1), f(x) approaches the value 2 (i.e., f(x)  2) we write this as lim ( ) 2 1   f x x It must be noted that it is not necessary for the function to be undefined at the point where limit is calculated. In the above example lim ( ) 2 f x x is the same as the value of function at x = 2 i.e., 3. Informally, we define limit as: Let f(x) be defined on an open interval about x0, except possibly at x0 itself. If f(x) gets arbitrarily close to L for all x sufficiently close to x0, we say that f approaches the limit L as x approaches x0, and we write f x L x x   lim ( ) 0 Sometimes, functions approach different values as x-approaches x0 from left and right. By left we mean x < x0 and right means x > x0. This is written as x   x0 and x   x0 respectively. e.g., f (x) = [x] (greatest integer function) For any integer n, lim ( )  1   f x n x n ...(i) and f x n x n    lim ( ) ...(ii) In such cases we say that lim f(x) xn does not exist. The limit in (i) is said to be the left hand limit (L.H.L.) at x = n and that in (ii) is called the right hand limit (R.H.L.) at x = n. LIMIT OF A FUNCTION 1 THEORY CONTENT OF LIMIT, CONTINUITY & DIFFERENTIABILITY DEFINITION 2
2.1 INFORMAL DEFINITION OF LIMITS Let f(x) be defined on an open interval about x0, except possibly at x0 itself. If f(x) gets arbitrarily close to L for all x sufficiently close to x0, we say that function approaches the limit L as x approaches x0, and we write f x L x x   lim ( ) 0 . This definition is “informal” because phrases like arbitrarily close and sufficiently close are imprecise, their meaning depends on the context. The definition is clear enough and enables us to recognize and evaluate limits of specific function. 2.2 FORMAL DEFINITION OF LIMIT Let f(x) be defined on an open interval about x0, except possible at x0 itself, we say that f(x) approaches the limit L as x approaches x0 and write f x L x x   lim ( ) 0 , if for every number  > 0, there exists a corresponding number  > 0 such that for all x 0 <  x  x0  <    f(x)  L  < . Illustration 1 Question: Show that limx1 (5x  3) = 2. Solution: Set x0 = 1, f(x) = 5x  3, and L = 2 in the definition of limit. For any given  > 0 we have to find a suitable  > 0 so that if x  1 and x is within distance  of x0 = 1, that is, if 0 <  x  1  < , then f(x) is within distance  of L = 2, that is  f(x)  2 <  We find  by working backwards from the  inequality  (5x  3)  2 =  5x  5 <  5 x  1 <   x  1 < (/5) Thus we can take  =  /5. If 0 <  x  1 <  =  /5, then x y y = 5x  3  3 0 2   2 2 +  5 1   5 1  1   (5x  3)2 =  5x  5| = 5 x  1 < 5( /5) =  This proves that limx1(5x  3) = 2. The value of  =  /5 is not the only value that will make 0 <  x  1 <  imply  5x  5 < . Any smaller positive  will do as well. The definition does not ask for a “best” positive , just one, that will work. For limit L to exist as x approaches x0, a function f must be defined on both sides of x0, and its values f(x) must approach L as x approaches x0 from either side. Because of this, ordinary limits are sometimes called two side limits. 2.3 RIGHT HAND LIMIT
If f x L x x    lim ( ) 0 for every number  > 0, there exists a corresponding number  > 0 such that for all x. x0 < x < x0 +    f(x)  L <  Then we call it right hand limit. e.g. 1 | | lim 0     x x x 2.4 LEFT HAND LIMIT If f x L x x    lim ( ) 0 for every number  > 0, there exists a corresponding number  > 0 such that for all x. x0   < x < x0   f(x)  L <  Then we call it left hand limit. e.g. 1 | | lim 0     x x x Through graph of | x | x we can easily visualize the things written above. y = 1 left hand limit right hand limit y = 1 Now, from the discussions we have just gone through we can easily say that limit of a function will exist iff. LHL and RHL both are finite, unique and equal. e.g. lim [ ] 1 x x will not exist as LHL = lim [ ] 0 1    x x ; RHL = lim [ ] 1 1    x x This can be seen graphically as LHL x 1 1 y RHL 2.5 ALGEBRA OF LIMITS
(i) 1 2 1 2 1 1 2 2 lim (c f(x) c g(x)) lim(c f(x)) lim(c g(x)) c l c l x a x a x a         , where c1 and c2 are given constants. (ii) 1 2 lim f(x). g(x) lim f(x). lim g(x) l . l x a x a x a      (iii) , 0 lim ( ) lim ( ) ( ) ( ) lim 2 2 1       l l l g x f x g x f x x a x a x a (iv) lim ( ( )) (lim ( )) ( ) 2 f g x f g x f l x a x a     , if and only if f(x) is continuous at x = l2. In particular, xa lim ln (g(x)) = ln l2 if l2 > 0. All these theorems must be used with utmost care. For example, we have assumed that l1 and l2 are finite. If these are not finite, the given theorems will not be applicable. e.g. 1 sin lim 0   x x x , and if we try to apply the theorems, we get x x x sin lim 0 = x x x x 1 lim sin . lim 0 0 , which does not exist. Which of course is an absurd result, we are getting this absurd result because in this case the given limit can not be written as the product of two limits as x x 1 lim 0 does not exist. Similarly,        x x x sin lim 0 , where [ . ] denotes the greatest integer function         x x x sin lim 0 . Here [x] is not continuous at x = 1. (v) ( ) log(1 ( )) lim ( ) 1 lim (1 ( )) g x f x g x x a x a f x e      2.6 EVALUATION OF LIMITS Following are indeterminate forms: (i) 0 0 (ii)   (iii) 0   (iv)    (v) 0° (vi) ° (vii) 1  We shall divide the ways of evaluation of limits in five categories: Simplification: In this method we can use (i) Direct substitution (ii) Rationalisation (iii) Factorization (iv) Use of formulas like binomial expansion, trigonometric formulas etc. (i) Direct substitution: We can directly substitute the number at which limit is to be find. e.g.  lim ( 3 2) 2 1    x x x can be find out by this method. lim ( 3 2) 2 2 1     x x x  lim | | 2 x x = 2 But before using this method, we have to see that LHL should remain equal to RHL.