Question :
Find the value of a and b in order that
$\displaystyle
\lim_{x \to 0} \frac{x(1 + a cosx) - b sinx}{x^3} = 1.
$
The numerator behaves like $\displaystyle x(1+a+b)-x^3(a/2 + b/6)$ for small $\displaystyle x$ and we need this to be $\displaystyle \sim x^3$ so $\displaystyle a=-5/2$ and $\displaystyle b=3/2$.
Note: you need to expand $\displaystyle x(1+ a \cos(x) )$ and $\displaystyle b \sin(x)$ up to and including the terms in $\displaystyle x^3$ then the coefficient of $\displaystyle x^3$ in the numerator should be $\displaystyle 1$ and the coefficients for all lower power terms should be $\displaystyle 0$ .
CB
You will find the required series for sin x and cos x here: Taylor series - Wikipedia, the free encyclopedia
Now apply the advice given in post #4.
$\displaystyle \lim_{x \to 0} \frac{x(1 - a cosx) - bsinx}{x^3} = 1$
$\displaystyle sinx = \sum_{n = 0}^{ \infty } \frac{(-1)^n}{(2n + 1)} = x - \frac{x^3}{3!}$
$\displaystyle cosx = \sum_{n = 0}^{ \infty} \frac{(-1)^n}{2n} x^{2n} = 1 - \frac{x^2}{2!}$
Substituting in the equation we get
$\displaystyle \lim_{x \to 0} \frac{x \left[ 1 + a \left(1 - \frac{x^2}{2!} \right) \right] - b \left(x - \frac{x^3}{3!} \right)}{x^3}$
$\displaystyle \lim_{x \to 0} \frac{x \left[ 1 + a \left( \frac{2 - x^2}{2} \right) \right] - b \left(\frac{6x - x^3}{6} \right)}{x^3}$
$\displaystyle \lim_{x \to 0} \frac{f(x)}{f(y)} = \lim_{x \to 0} \frac{f'(x)}{f'(y)} $
$\displaystyle \lim_{x \to 0} \frac{f'(x)}{f'(y)}= \lim_{x \to 0} \frac{ \frac{6x + 6ax - 3ax^3 - 6bx +bx^3}{6}}{3x^2} $
$\displaystyle \lim_{x \to 0} \frac{(1 + a - b) - \frac{1}{2} x^2(3a - b)}{3x^2}$
I am stuck here !!!!
What i have done is it correct or no?
The original limit is equal to (since we are ignoring a term or order $\displaystyle x$ in our truncation of the series):
$\displaystyle \lim_{x \to 0} \frac{x \left[ 1 + a \left( \frac{2 - x^2}{2} \right) \right] - b \left(\frac{6x - x^3}{6} \right)}{x^3} =$ $\displaystyle
\lim_{x \to 0} \frac{x(1+a-b)-x^3(a/2 +b/6)}{x^3}=$ $\displaystyle
\lim_{x \to 0} \frac{(1+a-b)}{x^2}- (a/2 +b/6)$
Now if this last limit is equal to $\displaystyle 1$, we must have $\displaystyle 1+a-b=0$ (as otherwise the limit would be infinite), and $\displaystyle a/2+b/6=-1$
CB
since $\displaystyle (1 + a - b) = 0$
therefore, $\displaystyle a = b - 1$
Substituting $\displaystyle a = b - 1$ in $\displaystyle 3a + b = -6$
$\displaystyle 3(b - 1) + b = -6$
$\displaystyle 3b - 3 + b = -6$
$\displaystyle 4b = -6 + 3$
$\displaystyle 4b = -3$
$\displaystyle b$ = - $\displaystyle \frac{3}{4}$
Substiting b in $\displaystyle 3a + b = -6$ weget
$\displaystyle 3a + \left( - \frac{3}{4} \right)$ = $\displaystyle -6$
$\displaystyle 3a$ = $\displaystyle -6 + \frac{3}{4}$
$\displaystyle 3a$ = $\displaystyle \frac{-24 + 3}{4}$
$\displaystyle 3a$ = $\displaystyle \frac{-21}{4}$
$\displaystyle a$ = $\displaystyle 3 \times \frac{-21}{4}$
a = - $\displaystyle \frac{63}{4}$
Is this correct ????
Does your answer satisfy
1 + a - b = 0 .... (1)
3a + b = -6 .... (2)
You should know how to solve simultaneous equations at the level you're apparently studying at. Personally, I'd just add equations (1) and (2) and solve for $\displaystyle a$. Then I'd substitute my value for $\displaystyle a$ into (1) and solve for $\displaystyle b$.
I am trying to solve the same question by using the L'Hospital rule , but i have got stuck at the following step
$\displaystyle \lim_{x \to 0} \ \frac{x(1+acosx)-bsinx}{x^3} = 1$
$\displaystyle f'(x) \ = \ x(1+acosx) - bsinx$
$\displaystyle = 1- ax (sinx) + (cosx) - b(cosx)$
and $\displaystyle g'(x) = 3 x^2$
$\displaystyle \lim_{x \to 0} \frac{ 1- ax (sinx) + (cosx) - b(cosx)}{3 x^2}$
$\displaystyle \lim_{x \to 0} \frac{1}{3x^2} - \frac{a}{3} \frac{sinx}{x} + (a-b). \frac{cosx}{3x^2}$
$\displaystyle \lim_{x \to 0} \frac{1}{3x^2} - \frac{a}{3} (1) + \lim_{x \to 0} (a-b). \frac{cosx}{3x^2}$
I am stuck here ............
Stop here. If this limit is finite the numerator must be equal to zero at $\displaystyle x=0$. Since otherwise the limit does not exist. So:
$\displaystyle 1+a-b=0$
or:
$\displaystyle b=1+a$
use this to substitute for $\displaystyle b$ in the last limit. Now another stage of L'Hopitals rule is needed
CB