Numerical Evaluation of a Limit

Perform all your calculations for this problem in double precision.

Consider the following limit:

$$\lim_{x \to 0} \frac{e^{x} - 1}{x}$$

Using L'Hopital's rule, it is straightforward to prove that the value of this limit is 1.

Write C++ code that numerically estimates the value of the above limit, by evaluating $\frac{e^{x} - 1}{x}$ for several values of x. In particular, use values of $x = 10^{-5}$, $10^{-6}$, $10^{-7}$, $10^{-8}$, $10^{-9}$, $10^{-10}$, $10^{-11}$, $10^{-12}$, $10^{-13}$, $10^{-14}$, and $10^{-15}$. Report your results at the end of this README.md file.

Now, we'll try a mathematically equivalent (but algorithmically different) approach to the same problem. We'll define $y = e^{x}$. We can then say that

$$\lim_{x \to 0} \frac{e^{x} - 1}{x} = \lim_{y \to 1} \frac{y - 1}{\ln{(y)}}$$

Write code that computes $y$ for the same set of values of $x$ as before ($x = 10^{-5}$, $10^{-6}$, $10^{-7}$, $10^{-8}$, $10^{-9}$, $10^{-10}$, $10^{-11}$, $10^{-12}$, $10^{-13}$, $10^{-14}$, and $10^{-15}$). Then evaluate $\frac{y - 1}{\ln{(y)}}$ for the resulting values of $y$. Remember that the log function from math.h computes the natural log of its argument. Report your results at the end of this README.md file.

At the end of this README.md file, explain your observations in detail, including the specific reasons for any differences in numerical stability.

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