Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to \pi /4}\frac{{\cot }^{3}x-\tan x}{\cos (x+\pi /4)}$ is :

Welcome, future engineer. Today, we stand before a limit that looks intimidating, but beneath its surface lies a beautiful, symmetric structure waiting to be revealed.

The problem is:

It is a classic JEE Advanced challenge that tests not just your knowledge of formulas, but your ability to see the 'soul' of an expression. Let us embark on this journey together.

The very first rule of limits is to plug in the value and check the form. If we substitute $x=\pi /4$ into the numerator, $\cot (\pi /4)$ is $1$, and $\tan (\pi /4)$ is also $1$. Thus, $1^3-1$ gives us $0$.

Now for the denominator, $\pi /4+\pi /4$ is $\pi /2$, and $\cos (\pi /2)$ is exactly $0$. We are dealing with a $0/0$ indeterminate form.

This is not a dead end; it is a signal. It means there is a hidden common factor in the numerator and denominator that is causing this zero, and our job is to hunt it down.

To reveal those hidden factors, our best strategy is to convert everything into $\sin x$ and $\cos x$. It is a classic move.

We know that $\cot x=\frac{\cos x}{\sin x}$, so ${\cot }^{3}x$ becomes $\frac{{\cos }^{3}x}{{\sin }^{3}x}$. Similarly, $\tan x=\frac{\sin x}{\cos x}$.

Let us rewrite our entire expression using these basic building blocks:

This might look more complex, but it is actually much more manageable.

Now, let us focus purely on the numerator. We have two fractions that we need to subtract. To do this, we take a common denominator, which will be ${\sin }^{3}x\cos x$.

Cross-multiplying the terms, the first term becomes ${\cos }^{4}x$, and the second term becomes ${\sin }^{4}x$. So, our new numerator is:

Take a moment to visualize this structure. It is a difference of squares, ${\cos }^{4}x-{\sin }^{4}x$, which can be factored as $({\cos }^{2}x-{\sin }^{2}x)({\cos }^{2}x+{\sin }^{2}x)$.

Since ${\cos }^{2}x+{\sin }^{2}x=1$, this simplifies beautifully to just ${\cos }^{2}x-{\sin }^{2}x$. And we can factor that again into $(\cos x-\sin x)(\cos x+\sin x)$.

Next, let us tackle the original denominator, $\cos (x+\pi /4)$. Using the compound angle formula, this expands to:

Do you see it? The term $(\cos x-\sin x)$ appears in both the numerator and the denominator! This was the culprit causing the $0/0$ form.

We can now safely cancel it out. What we are left with is:

With the problematic factor gone, we can finally substitute $x=\pi /4$ directly. The numerator becomes $\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=2$, and the denominator becomes $(\frac{1}{\sqrt{2}}{)}^4=1/4$.

Finally, $2/(1/4)=8$. You have navigated the complexity and arrived at the truth. The final answer is 8.