Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to \frac{\pi }{2}}\frac{\cot x-\cos x}{(\pi -2x{)}^3}$ equals:

Welcome, future engineer! Today, we are going to unravel a limit problem that might look like a tangled mess of trigonometry at first glance, but is actually a beautifully choreographed dance of algebraic simplification.

We want to evaluate the following limit:

The very first rule of the limit club is: always try direct substitution. If we plug in $x=\frac{\pi }{2}$, the numerator becomes $\cot (\frac{\pi }{2})-\cos (\frac{\pi }{2})=0-0=0$.

The denominator becomes $(\pi -2(\frac{\pi }{2}){)}^3=(\pi -\pi {)}^3=0$. We are staring at a $\frac{0}{0}$ indeterminate form.

This is not a dead end; it is an invitation to simplify.

Working with $x$ approaching $\frac{\pi }{2}$ is cumbersome. Let us make our lives easier by shifting the limit to zero.

We introduce a new variable, $t$, such that $x=\frac{\pi }{2}+t$. As $x\to \frac{\pi }{2}$, our new variable $t$ will naturally approach $0$.

This simple change of perspective is the compass that will guide us through the rest of the problem.

Now, let us transform our expression. Substituting $x=\frac{\pi }{2}+t$ into the numerator gives us $\cot (\frac{\pi }{2}+t)-\cos (\frac{\pi }{2}+t)$.

Using our trusty trigonometric identities, $\cot (\frac{\pi }{2}+t)=-\tan t$ and $\cos (\frac{\pi }{2}+t)=-\sin t$. The numerator becomes $-\tan t-(-\sin t)=\sin t-\tan t$.

Now for the denominator: $(\pi -2x{)}^3$ becomes $(\pi -2(\frac{\pi }{2}+t){)}^3=(\pi -\pi -2t{)}^3=(-2t{)}^3=-8t^3$.

Our limit now looks like this:

We can factor out a negative sign to make it:

We are almost there! Let us focus on $\frac{\tan t-\sin t}{t^3}$. By writing $\tan t$ as $\frac{\sin t}{\cos t}$, we get:

Now, we split this into three familiar pieces: $\frac{\sin t}{t}$, $\frac{1-\cos t}{t^2}$, and $\frac{1}{\cos t}$.

We know the following standard limits:

Combining these with our constant $\frac{1}{8}$, we get:

And there it is! A complex-looking expression reduced to a simple, elegant fraction. The final answer is $\frac{1}{16}$.