Evaluate the following using :L ' Hospital 's rule .
(i) if f(x) be a twice differentiable function and f'' (0) =2 , then find ` underset(x to 0) lim (2f(x)-3f(3x)+f(4x))/x^(2)`
(ii) if f(a) =2, f' (a) = 1, g (a) =2 , then find ` underset(x to 0) (g(x)f(a) -g(a)f(x0))/(x-a)`
(iii) ` underset( x to 0) (1/x - 1 /(sin x) )`
(iv) ` lim(x to 0+) x In x `
(v) ` underset(x to 0) | cot x|^( sin x) `
(vi) `underset(x to 0) (tan x + 4 tan 2x -3 tan 3x)/(x^(2) tan x) `

Explanation

We have
(i) ` underset(x to 0) lim (2f (x) -3f(2x) +f(4x))/x^(2) ` ,which in the form ` 0/0`
`underset(x to 0) lim(2f'(x)-6f'(2x)+4f'(4x))/(2x^(2)) ` using L' Hospital's rule , we differentiate N and `D^(r)` w.r.t x which is agian in the form `0/0`
`underset(x to 0)lim (2f''(x) -12f'''(2x)+4f'(4x))/2 `' Differentiate N and `D^(r)` w.r.t x.
` =( 2f''(0)-12f''(0)+16f''(0))/2`
` 3f''(0) = 3 xx 2= 6 " " f ''(0) =2`
(ii) we have
`f(a) =2, f'(a)=1, g(a) = -1 , g(a) =2`
`underset(x to a)lim (g (x) f(a) -g(a)f(x))/1 = g(a) f(a) -g(a) f(a)`
` 2 xx 2-1 xx -1 = 4 = 1 =5 `
(iii) ` underset( x to 0) lim(1/x - 1/(sin x) ) , ( oo - oo " form")`
Let us transform ` oo -oo " inot " 0/0` form and then we apply L : Hospital Rule.
`underset(x to 0)lim(1/x-1/sinx)=underset(x to 0) lim(sinx -x)/x^(2)(x/sinx)`
` = underset(x to 0)((sinx-x)/(x^(2))) underset(xto0) ( x/sinx) = underset(x to 0)lim (sin x -x)/x^(2) [0/0]`
`underset(x to 0) lim( cos x -1)/(2x) [0/0\] = underset(x to0) lim ( -sinx)/(2) =0`
(iv) we have ` underset(x to 0+)lim x x In x = underset(x to0+)lim("ln x")/1 [oo/oo] = underset(x to 0+)lim (1/x)/(-1/x^(2)) = underset(x to 0+)lim- (x)=0`
(v) We have `underset(xto 0)lim |cot x|^(sin x) =e^(underset(xto0)limsinx ln |cot x|)`
Now ` underset(x to 0) lim (sin x). ln |cot x| = underset(x to 0) lim (ln |cotx|)/(cosec x) [oo/oo]`
` = underset(x to0) lim (1/cotx(-cosec^(2)x0))/( -cosec x cot x) ` Differentiate `N^(r) and D^(r) ` w.r.t x.
`underset(x to 0+) lim (sinx)/(cos^(2)x) =0`
Hence the required limit = `e^(0)=1`
(vi) `underset(xto0)lim (tanx+ 4tan 2x - 3tan3x)/(x^(2)tanx) " " [0/0]`
`underset(xto0) lim(sec^(2)x +8 sec^(2) 2x -9 sec^(2)3x)/(2x tanx + x^(2) sec^(2)x) . " " [0/0]`
`= underset(x to 0)lim(2 sec^(2) xtan x+ 32 sec^(2) 2x tan 2x -54 sec^(2) 3x tan3x)/(2 tan x + 4x sec^(2) x + 2x^(2) sec^(2) x tan x) " " [0/0]`
`underset(x to 0)lim( 2sec^(4) x + 4sec^(2) x tan^(2) x + 64 sec^(4) 2x + 64 sec^(2) 2x tan^(2) 2x -162 sec^(4)3x - 324 sec^(2) 3x tan^(2) 3x)/(2 sec^(2) x + 4 sec^(2) x+ 8x sec^(2) x tan x + 4 x sec^(2) x tan x + 4x^(2) sec^(2) x tan^(2) x + 2x^(2) sec^(4) x)`
= ` ( 2 + 64 - 162)/(2+4) = -16`