Limits-Continuity-and-Differentiability-GATE-Study-Material-in-PDF-1.pdf

Cos x1 2x form again apply l hospital rule then 6 p a

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cos x−1 2x = ( 0 0 ) Form Again apply L Hospital Rule then
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6 | P a g e lim x→a f′′(x) g′′(x) = lim x→0 − sin x 2 = 0 Note: 1. If the functions are in the form of (0 × ∞) then reducing them into (or) 0 0 and then apply L’ hospital rule. 2. If the limit value is in the form of (∞ - ∞) then we need to follow this procedure to find the limit value. i. e. lim x→a (f(x) − g(x)) = lim x→a f(x) − lim x→a g(x) 3. If f(x) = ∞ and g(x) = ∞ as x a Then lim x→a [f(x) − g(x)] = lim x→a [ 1 g(x) 1 f(x) ] ÷ 1 f(x)g(x) Then apply L’ hospital form. Example 4: Find the value of lim x→ π 2 (sec x − tan x) Solution: lim x→ π 2 (sec x − tan x) = lim x→ π 2 ( 1 cos x sin x cosx ) = lim x→ π 2 (1−sinx) cos x Now it is in ( 0 0 ) form L − Hospital Rule gives = lim x→ π 2 − cos x − sinx = 0 Example 5: Find the value of lim x→0 x x
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7 | P a g e Solution: It is in the form of 0 0 and the following procedure is used to find the value of limit. Let y = x x log y = x log x lim x→0 log y = lim x→0 (x log x) lim x→0 log y = lim x→0 log x 1 x = lim x→0 1 x −1 x 2 = lim x→0 − x = 0 log y = 0 y = x x = e 0 = 1 Example 6: Find the value of lim x→ π 2 (tan x) cos x Solution: It is in the form of ( 0 ) Let y = tan x cos x ∴ lim x→ π 2 log y = lim x→ π 2 cos x log tan x = lim x→ π 2 logtan x sec x , Now, it is in the form of ( ) ∴ lim x→ π 2 1 tan x sec 2 x sec x tan x lim x→ π 2 sec x tan 2 x Again applying L- Hospital’s rule gives,
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8 | P a g e lim x→ π 2 sec x tan x 2 tan x sec 2 x = lim x→ π 2 cosx 2 = 0 y = e 0 = 1 Continuity A function f is said to be continuous at a number ‘a’ if lim x→a f(x) = f(a) Notice that above definition requires three things if f is constant at a: 1. f(x) is defined 2. lim x→a f(x) exist 3. lim x→0 f(x) = f(a) Example 7: For what value of k. the given function is continuous?
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