Limit

[man111]

[tex]\displaystyle\lim_{x\to 0} \frac{x(1+acosx)-bsinx}{(f(x))^3} = 1[/tex]. Then find [tex]a[/tex] and [tex]b[/tex]

Where [tex]f(x)[/tex] is a function of [tex]x[/tex]

[martin123456]

Since [tex]\lim_{x \to 0}(x(1+a\cos{x})-b\sin{x})=0[/tex] it must be true that [tex]\exist \lim_{x \to 0}f(x) = 0[/tex].
Then we use Lopital's rule. The limit is equivalent to [tex]\lim_{x \to 0}{\frac{1+a\cos{x}-xa\sin{x}-b\cos{x}}{3f^2(x)f'(x)}}=1[/tex]. If [tex]\lim_{x \to 0}{3f^2(x)f'(x)} \ne 0[/tex] then we can't solve the problem. So we consider [tex]\lim_{x \to 0}{3f^2(x)f'(x)} = 0[/tex]. Then it's a must [tex]\lim_{x \to 0}{(1+a\cos{x}-xa\sin{x}-b\cos{x})}=0[/tex] and so [tex]1+a-b=0[/tex]. Again Lopital's rule [tex]\lim_{x \to 0}{\frac{-2a\sin{x}-xa\cos{x}+b\sin{x}}{6f(x)f'(x)^2+3f^2(x)f''(x)}}=1[/tex]. Since [tex]\lim_{x \to 0}{(-2a\sin{x}-xa\cos{x}+b\sin{x})} = 0[/tex] we must have that [tex]\lim_{x \to 0}{(6f(x)f'(x)^2+3f^2(x)f''(x) )= 0[/tex]. Then again Lopitals' rule to get [tex]\lim_{x \to 0}{\frac{-3a\cos{x}+xa\sin{x}+b\cos{x}}{6f'(x)^3+6f(x)2f'(x)f''(x)+6f(x)f'(x)f''(x)+3f^2(x)f''(x)}} = 1[/tex]. Again assume that the denominator is 0, then the numerator must be zero to when x tends to 0. So we get [tex]-3a+b=0[/tex].

So a lot of assumptions here to get an answer of [tex](\frac{1}{2},\frac{3}{2})[/tex] for a functoion [tex]f[/tex] that satisfies [tex]\exist f'(0),f''(0)[/tex], [tex]\lim_{x \to 0}f'(x) \ne \infty[/tex], and so on

[man111]

martin123456 thanks for brilliant efford.

actually here [tex]x^3[/tex] instead of [tex](f(x))^3[/tex]

[mkmarinov]

pwnt