Evaluate $\lim_{x \rightarrow 0}\left(\frac{(1+57x)^{67}-(1+67x)^{57}}{x^{2}} \right)$. Calculate:
$$\lim_{x \rightarrow 0}\left(\frac{(1+57x)^{67}-(1+67x)^{57}}{x^{2}} \right)$$
Without using L'Hospital's rule.
 A: Recall the Taylor expansion (which is also a binomial expansion, if you prefer): $(1+u)^\alpha=1+\alpha u+\binom{\alpha}{2}u^2+O(u^3)$. Then
$$
f(x)=(1+57x)^{67}=1+3819x+7183539x^2+O(x^3)
$$
and
$$
g(x)=(1+67x)^{57}=1+3819x+7164444x^2+O(x^3).
$$
So
$$
\frac{f(x)-g(x)}{x^2}=\frac{19095x^2+O(x^3)}{x^2}=19095+O(x)\longrightarrow 19095.
$$
A: Hint: Use the binomial expansion. After a few terms, things will cancel out or go to $0$ in the limit and you'll be left with the relevant value.
A: The binomial formula gives:
$$(1+57x)^{67} =1+67(57x)+2211(57x)^2 +47905(57x)^3 + \cdots+(57x)^{67}$$
$$(1+67x)^{57}=1+57(67x)+1596(67x)^2+29260(67x)^3+...+(67x)^{57}$$
$$\implies (1+57x)^{67}-(1+57x)^{67}=(2211\times 57^2-1596 \times 67^2)x^2+ p(x)x^3$$
where $p$ is a polynomial in $x$.
The $x^2$ term is $(2211\times 57^2-1596\times 67^2)=19095$, and since $\lim_{x \to 0} \frac{p(x) x^3}{x^2} =0$, the desired value of the limit is $19095$.
A: In general,
$$
\begin{align}
&\lim_{x\to0}\left(\frac{(1+mx)^n-(1+nx)^m}{x^2}\right)\\
&=\small\lim_{x\to0}\left(\frac{\left(1+\frac{n}{1}(mx)+\frac{n(n-1)}{1\cdot2}(mx)^2+O(x^3)\right)
-\left(1+\frac{m}{1}(nx)+\frac{m(m-1)}{1\cdot2}(nx)^2+O(x^3)\right)}{x^2}\right)\\
&=\frac{n(n-1)}{1\cdot2}m^2-\frac{m(m-1)}{1\cdot2}n^2\\[14pt]
&=\frac{(n-m)nm}{2}
\end{align}
$$
In particular,
$$
\frac{(67-57)67\cdot57}{2}=19095
$$
A: $$\lim_{x \rightarrow 0}\left(\frac{(1+mx)^{n}-(1+nx)^{m}}{x^{2}} \right) = \frac{(n-m)mn}{2}$$
Proof:
$=\lim_{x \rightarrow 0}\left(\frac{\left({n\choose0}(mx)^0+{n\choose1}(mx)^1+{n\choose2}(mx)^2+{n\choose3}(mx)^3+\:\cdots\:+{n\choose n}(mx)^n\right) \;-\; \left({m\choose0}(nx)^0+{m\choose1}(nx)^1+{m\choose2}(nx)^2+{m\choose3}(nx)^3+\:\cdots\:+{m\choose m}(nx)^m\right)}{x^{2}} \right)$
$=\lim_{x \rightarrow 0}\left(\frac{1+nmx+{n\choose2}m^2x^2+{n\choose3}m^3x^3+\:\cdots\:+m^nx^n - 1-nmx+{m\choose2}n^2x^2-{m\choose3}n^3x^3-\:\cdots\:-\,n^mx^m}{x^{2}} \right)$
$=\lim_{x \rightarrow 0}\left(\frac{{n\choose2}m^2x^2+{n\choose3}m^3x^3+\:\cdots\:+m^nx^n - {m\choose2}n^2x^2-{m\choose3}n^3x^3-\:\cdots\:-n^mx^m}{x^{2}} \right)$
Canceling $x^2$ from numerator and denominator:
$=\lim_{x \rightarrow 0}\left({n\choose2}m^2+{n\choose3}m^3x+\:\cdots\:+m^nx^{n-2} - {m\choose2}n^2-{m\choose3}n^3x-\:\cdots\:-n^mx^{m-2} \right)$
$={n\choose2}m^2 - {m\choose2}n^2$
$=\frac{n(n-1)}{2}m^2 - \frac{m(m-1)}{2}n^2$
$=\frac{(n-m)mn}{2}$
