这是一份BATH巴斯大学ME10305作业代写的成功案例
$f^{\prime}(x)$ and $g^{\prime}(x)$ exist near, but not necessarily at, $a$. Then, if
$$
f^{\prime}(x) / g^{\prime}(x)
$$
tends to a limit $\lambda$ as $x \rightarrow a$, the limit of $f(x) / g(x)$ exists and equals $\lambda$ also.
By Cauchy’s formula, if $a<z_{1}<x$,
$$
\frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f^{\prime}\left(z_{1}\right)}{g^{\prime}\left(z_{1}\right)}
$$
Hence, assuming the limit in question exists,
$$
\lim {x \rightarrow a+0} \frac{f(x)}{g(x)}=\lim {z_{1} \rightarrow a+0} \frac{f^{\prime}\left(z_{1}\right)}{g^{\prime}\left(z_{1}\right)}=\lambda
$$
Here it has been assumed that $x>a$ and so only the right-hand limits have been taken-the point $a$ being approached from above.
If $x<a$, that is if $x<z_{2}<a$, a similar argument yields
$$
\lim {x \rightarrow a-0} \frac{f(x)}{g(x)}=\lim {z_{x} \rightarrow a-0} \frac{f^{\prime}\left(z_{2}\right)}{g^{\prime}\left(z_{2}\right)}=\lambda
$$
where left-hand limits have been taken.
ME10305 COURSE NOTES ：
Complex numbers were introduced first by Cardan in his examination of the solutions of cubic equations. Hence, consider the cubic equation $x^{3}-1=0$. This may be written
$$
(x-1)\left(x^{2}+x+1\right)=0 .
$$
Hence, the roots of the original cubic equation are $x=1$ and the roots of the quadratic equation
$$
x^{2}+x+1=0
$$
and the roots of this equation are
$$
x=1 / 2(-1 \pm V-3)
$$