Log approximation Can somebody help me with this approximation:

as $m$ gets really large, $\dfrac{r}{m}$ gets really small and hence $\log\left(1+\dfrac{r}{m}\right) \sim \dfrac{r}{m}$.

I don't understand how did $\log\left(1+\dfrac{r}{m}\right)$ approximately equals to $\dfrac{r}{m}$? Please clarify. 
 A: The natural logarithm of $1 + r/m$ is defined to be the area under the curve $t\to 1/(1+t)$ between $0$ and $r/m$.

When $r/m$ is small, this region (shown in blue) is approximated moderately well by a rectangle of height $1$ and base $r/m$ (shown in gray underneath), whose area is $r/m$. The relative error in the approximation is no greater than the decrease in the value of $1/(1+t)$ over the region, which for small $t$ becomes arbitrarily close to zero.  That answers the question.

From the figure it is clear that the region is much better approximated by a trapezoid with vertices at the origin, $(x, 0)$, $(x, 1/(1+x))$, and $(0,1)$, where $x=r/m$. Its area is
$$x\left(1 + \frac{1}{1 +x}\right)/2 = x\left(1 - \frac{x}{2+2x}\right).$$
In the next figure, these two approximations are plotted (in red for the trapezoid, gold for the rectangle) along with a graph of $\log(1+x)$ (in blue) for smallish values of $x$.  There is scarcely any visible difference between the trapezoidal approximation and the logarithm. All three values converge to one another as $x$ approaches $0$.

A: Shortly, consider the Taylor expansion of your term:
$$\log(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} \pm \ldots.$$
Now, if $x$ (which stands here for your fraction $\frac{r}{m}$) becomes small, you can well approximate your function by the linear term in the Taylor series.
Edit: I see this already mentioned in the comments to the original question... so I won't claim priority for this deep result ;-)
