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Calculate:

$$\lim_{x \rightarrow 0}\left(\frac{(1+57x)^{67}-(1+67x)^{57}}{x^{2}} \right)$$

Without using L'Hospital rule

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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, $$ \frac{(67-57)67\cdot57}{2}=19095 $$

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  • $\begingroup$ One of these cases where the generalization is much nicer to read than the particular case, +1. $\endgroup$ – Julien Mar 21 '13 at 14:47
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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 $$ \frac{f(x)-g(x)}{x^2}=\frac{19095x^2+O(x^3)}{x^2}=19095+O(x)\longrightarrow 19095. $$

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  • $\begingroup$ Just wondering, notation, is $O(x)$ and $\mathcal{O}(x)$ same thing ? $\endgroup$ – Cortizol Mar 21 '13 at 16:30
  • $\begingroup$ @Cortizol I guess so. I meant Laudau big O. $\endgroup$ – Julien Mar 21 '13 at 16:39
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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.

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la formule de binôme donne (1+57*x)^67=1+67*(57*x)+2211*(57*x)^2+47905*(57*x)^3+...+(57*x)^67

(1+67*x)^57=1+57*(67*x)+1596*(67*x)^2+29260*(67*x)^3+...+(67*x)^57

(1+57*x)^67-(1+57*x)^67=(2211*57^2-1596*67^2)*x^2+p(x)*x^3

avec p un polynome en x

or (2211*57^2-1596*67^2)=19095

et lim (p(x)*x^3) /x^2 =0 (quand x tend vers 0)

donc la limite cherchée est 19095.

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$$\lim_{x \rightarrow 0}\left(\frac{(1+mx)^{n}-(1+nx)^{m}}{x^{2}} \right) = \frac{(n-m)mn}{2}$$

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:

$=\lim_{x \rightarrow 0}\left({n\choose2}m^2+{n\choose3}m^3x+\:\cdots\:+m^nx^{n-2} - {m\choose2}n^2-{m\choose3}n^3x-\:\cdots\:-n^mx^{m-2} \right)$

$={n\choose2}m^2 - {m\choose2}n^2$

$=\frac{n(n-1)}{2}m^2 - \frac{m(m-1)}{2}n^2$

$=\frac{(n-m)mn}{2}$

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