Limits with L'Hôpital's rule

Question

Find the values of $a$ and $b$ if $$ \lim_{x\to0} \dfrac{x(1+a \cos(x))-b \sin(x)}{x^3} = 1 $$ I think i should use L'Hôpital's rule but it did not work.

Answer 1

The easiest way (in my opinion) is to plug in the power series expansion of $x(1+a\cos x)-b\sin x$ around zero. Then, the limit becomes

$$\lim_{x\rightarrow 0} \frac{x(a-b+1)+x^3(b-3a)/6+O(x^5)}{x^3}=1.$$

Now you have two equations involving $a$ and $b$ to satisfy (can you figure out what those equations are?)

Answer 2

use the equivalent, near $0$ \begin{eqnarray*} \cos x &\approx &1-\frac{x^{2}}{2} \\ \sin x &\approx &x-\frac{x^{3}}{6} \end{eqnarray*} \begin{eqnarray*} \frac{x(1+a\cos x)-b\sin x}{x^{3}} &\approx &\frac{x(1+a\left( 1-\frac{x^{2}% }{2}\right) )-b\left( x-\frac{x^{3}}{6}\right) }{x^{3}} \\ &=&\frac{x+ax-a\frac{x^{3}}{2}-bx+\frac{bx^{3}}{6}}{x^{3}} \\ &=&\frac{(1+a-b)x+x^{3}(\frac{b-3a}{6})}{x^{3}} \\ &=&\frac{(1+a-b)}{x^{2}}+\frac{b-3a}{6} \end{eqnarray*} It suffices to choose $a$ and $b$ such that \begin{equation*} (1+a-b)=0\ \ \ \ \ and\ \ \ \ \ \ b-3a=6 \end{equation*} that is \begin{equation*} a=-\frac{5}{2},\ \ \ and\ \ b=-\frac{3}{2} \end{equation*}

Answer 3

You can, unless using L'Hospital's rule repeatedly, which is not the alpha and omega of limit computations. The simplest way to go is to use Taylor's polynomial:

$$ \cos x=1-\dfrac{x^2}2+o(x^2),\quad \sin x=x-\frac{x^3}6+o(x^3),$$ whence $$x(1+a\cos x)-b\sin x=(1+a-b)x+\Bigl(\frac b6-\frac a2\Bigr)x^3+o(x^3) $$ $$\frac{x(1+a\cos x)-b\sin x}{x^3}= \frac{(1+a-b)x+\Bigl(\dfrac b6-\dfrac a2\Bigr)x^3+o(x^3)}{x^3}. $$ For the limit to be equal to $1$, the following equations must be satisfied: $$\begin{cases} a-b+1=0,\\\dfrac b6-\dfrac a2=1. \end{cases}$$