Limits of Functions

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  • 1. LIMIT OF FUNCTIONS
  • 2. DEFINITION Limit of a function f(x) is said to exist as, x  a when, f (a - h) = f (a + h) = some finite value M. Limit Limit   h 0 h 0 (Left hand limit) (Right hand limit) Note that we are not interested in knowing about what happens at x = a. Also note that if L.H.L. & R.H.L. are both tending towards  or ‘–’ then it is said to be infinite limit. Remember, xa Limit x a
  • 3. EXAMPLES
  • 4. INDETERMINANT FORMS  0 ,  , 0  ,  - , º, 0º, and 1. 0   Note :  0 doesnt means exact zero but represent a value approaching towards zero similary to 1 and infinity.  +=  x=  (a/) = 0 if a is finite a  is not defined for any a  R. 0  a b = 0, if & only if a = 0 or b = 0 and a & b are finite.
  • 5. EXAMPLES
  • 6. METHOD OF REMOVING INDETERMINANCY To evaluate a limit, we must always put the value where x is approaching to in the function. If we get a determinate form, then that value becomes the limit otherwise if an indeterminant form comes. Then apply one of the following methods: Factorisation Rationalisation or double rationalisation Substitution Using standard limits Expansions of functions.
  • 7.  Factorization method We can cancel out the factors which are leading to indeterminacy and find the limit of the remaining expression. Rationalization /Double Rationalization We can rationalize the irrational expression by multiplying with their conjugates to remove the indeterminacy.
  • 8. EXAMPLES
  • 9. FUNDAMENTAL THEOREMS ON LIMITS Let Limit f (x) = l & Limitg (x) = m. If & m exists then x a x a  Limit { f (x) ± g (x) } = ± m x a  Limit { f(x). g(x) } = l. m x a f ( x)   Limit x a g ( x ) = , provided m 0 m  Limit k f(x) = k f(x) ; where k is a constant. x a    Limit f [g(x)] = f  Limitg ( x ) = f (m); provided f is continuous  x a  x a   at g (x) = m.
  • 10. EXAMPLES
  • 11. STANDARD LIMITS Sinx tanx tan 1 x Sin 1 x Lim  1  Lim  Lim  Lim x 0 x x 0 x x 0 x x 0 x [ Where x is measured in radians ] x Lim (1 + x)1/x = e ; x 0 Lim 1   = e  1 x   x ex 1 a x 1 Lim  1; Lim  log e a, a  0 x 0 x 0 x x ln(1  x) Lim 1 x 0 x xn  an n 1 Lim  na x a xa
  • 12. EXAMPLES
  • 13. USE OF SUBSITUTION IN SOLVING LIMITPROBLEMS Sometimes in solving limit problem we convert Lim f(x) x a by substituting x = a + h or x = a – h as Lim f(a + h) or Lim f(a – h) according as need of the h 0 h 0 problem.LIMIT WHEN X   Since x    1/x  0 hence in this type of problem we express most of the part of expression in terms of 1/x and apply 1/x  0.
  • 14. EXAMPLES
  • 15. LIMITS USING EXPANSION x ln a x 2 ln2 a x 3 ln3 a a 1 x    .........  0 a 1! 2! 3! x x2 x3 e x 1     ...... 1! 2! 3! x2 x3 x4 ln(1+x) = x    .........  1  x  1 for 2 3 4 x3 x5 x7 sin x x     ..... 3! 5! 7! x2 x4 x6 cosx 1     ..... 2! 4! 6! x 3 2x 5 tan x = x    ...... 3 15
  • 16. x3 x5 x7 tan-1x = x     .... 3 5 7 12 3 12.3 2 5 12.3 2.5 2 7 sin-1x = x  x  x  x  ..... 3! 5! 7! x 2 5x 4 61x 6 sec-1x = 1     ...... 2! 4! 6! for |x| < 1, n  R n(n  1) 2 n(n  1)(n  2) 3(1  x)  1  nx  n x  x  ......... 1.2 1.2.3
  • 17. EXAMPLES
  • 18. LIMITS OF FORM 1, 00, 0 0 All these forms can be convered into form in the following ways 0 1. If x  1, y   , then z = (x)y  ln z = y ln x  ln z = nx (1/ y ) Since y   hence 1/y  0 and x  1 hence lnx  0 2. If x  0, y  0, then z = x y  ln z = y ln x y 0  ln z = 1/ ny = form 0
  • 19. 3. If x   , y  0, then z = x y  ln z = y ln x y  ln z = = 0 form 1/ nx 0
  • 20.  For (1) type of problems we can use following rules  lim (1 + x)1/x = e x 0  lim [f(x)]g(x) x a where f(x)  1 ; g(x)   as x  a 1 { f ( x ) 1} . g( x ) 1  f ( x)  1 f ( x ) 1 lim = x a lim [ f ( x )1] g( x ) x a = e
  • 21. EXAMPLES
  • 22. SANDWICH THEOREM OR SQUEEZE PLAYTHEOREM If f(x)  g(x)  h(x) " x Lim f ( x)  l  Lim h( x) xa xa then Lim g ( x)  l xa
  • 23. SOME IMPORTANT NOTES ln x x1. lim 0 2. lim x  0 x  x x  e As x  , ln x increases much slower than any (+ve) power of x where ex increases much faster than (+ve) power of x.3. lim(1  h)n  0 & lim(1  h)   , where h > 0. n n  n4. If limf(x) = A > 0 & lim (x) = B (a finite quantity) f xa xa  ( x) then; lim[ f ( x)] e z where x a z  lim  ( x).ln[ f ( x)]  e B ln A A B x a
  • 24. EXAMPLES