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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, xa 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 xa
- 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) xa xa then Lim g ( x) l xa
- 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 n4. If limf(x) = A > 0 & lim (x) = B (a finite quantity) f xa xa ( 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

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