Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of ${\lim }_{h\to 0}\left\{\frac{\sqrt{3}\sin \left(\frac{\pi }{6}+h\right)-\cos \left(\frac{\pi }{6}+h\right)}{\sqrt{3}h\left(\sqrt{3}\cosh -\sinh \right)}\right\}$ is:

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of trigonometric functions. You are staring at the limit:

When you first see this, your instinct might be to panic. It looks like a tangled web of sines, cosines, and fractions. But remember the golden rule of limits: always check the form first.

If you substitute $h=0$, the numerator becomes $\sqrt{3}\sin (\frac{\pi }{6})-\cos (\frac{\pi }{6})=\sqrt{3}(\frac{1}{2})-\frac{\sqrt{3}}{2}=0$. The denominator also collapses to zero. We have a $\frac{0}{0}$ indeterminate form.

This is not a wall; it is a door. It tells us that there is a hidden factor of $h$ waiting to be cancelled out.

The Art of Transformation

Look at the numerator: $\sqrt{3}\sin (\frac{\pi }{6}+h)-\cos (\frac{\pi }{6}+h)$. This is a classic pattern: $a\sin \theta -b\cos \theta$.

In the world of JEE Advanced, whenever you see this, you must immediately think of the harmonic addition theorem. We need to factor out $\sqrt{a^2+b^2}$. Here, $a=\sqrt{3}$ and $b=1$, so $\sqrt{(\sqrt{3}{)}^2+1^2}=2$.

By factoring out $2$, we get $2[\frac{\sqrt{3}}{2}\sin (\frac{\pi }{6}+h)-\frac{1}{2}\cos (\frac{\pi }{6}+h)]$. Now, look closely at those fractions. $\frac{\sqrt{3}}{2}$ is $\cos (\frac{\pi }{6})$ and $\frac{1}{2}$ is $\sin (\frac{\pi }{6})$.

The expression becomes $2[\cos (\frac{\pi }{6})\sin (\frac{\pi }{6}+h)-\sin (\frac{\pi }{6})\cos (\frac{\pi }{6}+h)]$. This is the expansion of $\sin (A-B)$ where $A=\frac{\pi }{6}+h$ and $B=\frac{\pi }{6}$. The numerator simplifies beautifully to $2\sin (h)$.

The Denominator's Secret

Now, let's turn our attention to the denominator. We have $\sqrt{3}\cos h-\sin h$. It is the same pattern!

We factor out $2$ again, giving us $2[\frac{\sqrt{3}}{2}\cos h-\frac{1}{2}\sin h]$. Substituting the same trigonometric values, we get $2[\cos (\frac{\pi }{6})\cos h-\sin (\frac{\pi }{6})\sin h]$.

This is the identity for $\cos (A+B)$. Thus, the inner part of the denominator becomes $2\cos (h+\frac{\pi }{6})$.

The Climax

We have now stripped away the complexity. Let's reassemble our limit with these simplified parts:

The factor of $2$ in the numerator and denominator cancels out, leaving us with:

We know the fundamental limit ${\lim }_{h\to 0}\frac{\sin h}{h}=1$. As $h$ approaches $0$, the remaining term $\cos (h+\frac{\pi }{6})$ approaches $\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}$.

Substituting these values, we get:

Look at what we achieved. We took a terrifying expression and, through the elegance of trigonometric identities, reduced it to a simple constant.

This is the beauty of mathematics. It is not about memorizing formulas; it is about recognizing patterns and having the patience to peel back the layers.

You have mastered the harmonic addition trick today—keep this in your toolkit, for it will serve you well in the exam hall. Keep practicing, keep questioning, and keep pushing forward.