Just Check your limits

Taking the natural log of both sides of the given equation, we then have the equation

$n = \displaystyle\lim_{x \to a} \tan(\frac{\pi x}{2a}) \ln(2 - \frac{a}{x}) = \lim_{x \to a} \dfrac{\ln(2 - \frac{a}{x})}{\cot(\frac{\pi x}{2a})}$ .

This limit is of the indeterminate form $\frac{0}{0}$ , so applying L'Hopital's rule we have that

$n = \displaystyle\lim_{x \to a} \dfrac{\dfrac{\frac{a}{x^{2}}}{2 - \frac{a}{x}}}{-\frac{\pi}{2}\csc^{2}(\frac{\pi x}{2a})} = -\dfrac{2}{\pi} * \lim_{x \to a} \dfrac{\dfrac{a}{2x - a}}{\csc^{2}(\frac{\pi x}{2a})} = -\dfrac{2}{\pi} * \dfrac{\frac{a}{a}}{\csc^{2}(\frac{\pi}{2})} = -\dfrac{2}{\pi} = \boxed{-0.6366}$

to 4 decimal places.

Just apply standard method of solving limits of sort 1^infinity. Lim(x-->a){1+f(x)}^1/g(x)= e^[lim(x-->a){f(x)/g(x)}]

My solution is quite long but I think it works.

We already know that $e^x=\displaystyle\lim_{n \to \infty} (1+\frac{x}{n})^{n}$ . So we would like to transform this.

Let $y=\tan(\frac{\pi}{2} \frac{x}{a})$ and so $\frac{a}{x}=\frac{\pi}{2}\cdot \frac{1}{\arctan(y)}$

When $x$ approaches $a$ , $y$ approaches $\infty$

Then we need to compute $\displaystyle\lim_{y \to \infty} (1+\frac{\alpha}{y})^{y}$

where $\alpha = y-\frac{\pi}{2}\cdot \frac{y}{\arctan(y)}$

So because $e^x=\displaystyle\lim_{n \to \infty} (1+\frac{x}{n})^{n}$ , this means that the answer $n$ equals :

$\displaystyle\lim_{y \to \infty} y-\frac{\pi}{2}\cdot \frac{y}{\arctan(y)} \iff \displaystyle\lim_{y \to \infty} \frac{1}{\arctan(y)}(y\arctan(y)-\frac{\pi}{2}\cdot y)$

$\iff \displaystyle\lim_{y \to \infty} \frac{1}{\arctan(y)}(y \cdot (\arctan(y)-\frac{\pi}{2}))$

However $\displaystyle\lim_{y \to \infty} y \cdot (\arctan(y)-\frac{\pi}{2}) \iff \displaystyle\lim_{y \to \infty} \frac{\arctan(y)-\frac{\pi}{2}}{\frac{1}{y}}$ .

This limit is of the indeterminate form $\frac{0}{0}$ , so by L'Hopital's rule

$\displaystyle\lim_{y \to \infty} \frac{\arctan(y)-\frac{\pi}{2}}{\frac{1}{y}}=\displaystyle\lim_{y \to \infty} -\frac{y^2}{y^2+1}=-1$ .

Thus, we get that $n=\displaystyle\lim_{y \to \infty} \frac{1}{\arctan(y)}(-1)=-\frac{2}{\pi}$ as $\displaystyle\lim_{x \to \infty} \arctan(x)=\frac{\pi}{2}$