Q.Evaluate: $\int \dfrac{e^{6\log x}-e^{5\log x}}{e^{4\log x}-e^{3\log x}}\,dx$

We are asked to compute

where

The key idea: both numerator and denominator vanish as $x \to 0$, so we need series expansions (or L’Hôpital’s rule) to find the leading behavior. Since the difference may cancel lower-order terms, we expand each function up to the constant term.

1. Expand $f(x)$ near $x=0$
   • $a^x = e^{x \log a} = 1 + x \log a + \frac{x^2 (\log a)^2}{2} + O(x^3)$.
   • So $a^x - 1 = x \log a + \frac{x^2 (\log a)^2}{2} + O(x^3)$.
   • Numerator: $x(a^x - 1) = x^2 \log a + \frac{x^3 (\log a)^2}{2} + O(x^4)$.
   • Denominator: $1 - \cos x = \frac{x^2}{2} - \frac{x^4}{24} + O(x^6)$.
   • Hence

• For small $x$, the denominator tends to $1/2$, so

2. Expand $g(x)$ near $x=0$
   • First, the denominator factor:

 So  

• Also $a^x = 1 + x \log a + \frac{x^2 (\log a)^2}{2} + O(x^3)$, so $1 - a^x = -x \log a - \frac{x^2 (\log a)^2}{2} + O(x^3)$.
• The numerator of $g$: $x(1 - a^x) = -x^2 \log a - \frac{x^3 (\log a)^2}{2} + O(x^4)$.
• The denominator of $g$: $a^x (\sqrt{1 - x^2} - \sqrt{1 + x^2}) = (1 + x \log a + \cdots)(-x^2 + O(x^4)) = -x^2 - x^3 \log a + O(x^4)$.
• Therefore $$g(x) = \frac{-x^2 \log a - \frac{x^3 (\log a)^2}{2} + \cdots}{-x^2 - x^3 \log a + \cdots} …$$

The key insight is that the expression inside the limit is a disguised Riemann sum. When you see a sum of the form $\frac{1}{n} \sum f(k/n)$, you think of an integral. Here, the terms are $k e^{k/n}$ but the sum runs up to $2n$, not $n$, so the interval is $[0, 2]$. The factor $\frac{1}{n^2}$ is actually $\frac{1}{n} \cdot \frac{1}{n}$, where one $\frac{1}{n}$ is the width of subintervals and the other $\frac{1}{n}$ is part of the function value $k/n$ times $e^{k/n}$. Let's unpack this carefully.

1. Rewrite the sum in Riemann-sum form. The general term is $k e^{k/n}$. Notice that $k = n \cdot (k/n)$, so

Then the whole sum becomes

2. Identify the Riemann sum.

   The expression $\frac{1}{n} \sum_{k=1}^{2n} f\!\left(\frac{k}{n}\right)$ with $f(x) = x e^x$ is a right-endpoint Riemann sum for $\int_0^2 f(x)\,dx$, because the points $x_k = k/n$ range from $1/n$ to $2n/n = 2$, and the subinterval width is $\Delta x = 1/n$. As $n \to \infty$, this sum converges to the integral.

3. Compute the integral.

Use integration by parts: let $u = x$, $dv = e^x dx$, so $du = dx$, $v = e^x$. Then

Evaluate from $0$ to $2$:

The question asks what happens when you compound interest infinitely many times per year. The expression $P \left( 1 + \frac{r}{100n} \right)^{tn}$ is the familiar compound interest formula where $P$ is principal, $r$ is the annual interest rate in percent, $t$ is the number of years, and $n$ is the number of compounding periods per year. As $n \to \infty$, you are compounding every instant — this is called continuous compounding.

The key idea is that the limit $\lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x$ is one of the most famous limits in mathematics. Here, the exponent is $tn$, not just $n$, so we need to adjust the form to match that standard limit.

Let’s work through it step by step.

1. Rewrite the expression to isolate the standard limit form.

   We have $P \left( 1 + \frac{r}{100n} \right)^{tn}$. Let $x = \frac{r}{100}$. Then the expression becomes $P \left( 1 + \frac{x}{n} \right)^{tn}$.

2. Manipulate the exponent.

   Write $(1 + \frac{x}{n})^{tn} = \left[ \left(1 + \frac{x}{n}\right)^n \right]^t$. This is valid because $(a^m)^k = a^{mk}$. So the limit is $P \cdot \left[ \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right]^t$.

3. Apply the standard limit. …