Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f:\mathbb{R}\to \mathbb{R}$ be a function given by $f(x)=\{\begin{array}{ll}\frac{1-\cos 2x}{x^2},& x<0\\ \alpha ,& x=0\\ \frac{\beta \sqrt{1-\cos x}}{x},& x>0\end{array}$ where $\alpha ,\beta \in \mathbb{R}$. If $f$ is continuous at $x=0$, then ${\alpha }^2+{\beta }^2$ is equal to :

Imagine you are an engineer tasked with building a bridge that connects two roads. For the bridge to be safe and perfectly smooth, the left road, the right road, and the bridge itself must all meet at the exact same point.

In the world of calculus, this is the essence of continuity. We are given a function $f(x)$ defined in three pieces, and our goal is to ensure that these pieces meet seamlessly at $x=0$.

For $f(x)$ to be continuous at $x=0$, the left-hand limit, the right-hand limit, and the function value at $x=0$ must all be equal. We know $f(0)=\alpha$, so our mission is to find $\alpha$ and $\beta$ such that the limits from both sides equal $\alpha$.

Unmasking the Indeterminate

Let us analyze the left side of our bridge, where $x<0$. We are looking at the limit:

If we try to plug in $x=0$ directly, we get $1-\cos (0)=0$ in the numerator and $0^2=0$ in the denominator. This is the classic $\frac{0}{0}$ indeterminate form.

We use the trigonometric identity $1-\cos 2\theta =2{\sin }^2\theta$. Substituting this into our limit, the expression transforms into:

Since we know the standard limit ${\lim }_{x\to 0}\frac{\sin x}{x}=1$, the LHL becomes $2(1{)}^2=2$. Because the function is continuous, this LHL must equal $f(0)$, which is $\alpha$. Thus, we have found our first piece of the puzzle: $\alpha =2$.

Navigating the Square Root

Now, let us turn to the right side of the bridge, where $x>0$. We are dealing with the limit:

We apply the half-angle identity $1-\cos x=2{\sin }^2(x/2)$. The expression becomes:

Since $x\to 0^{+}$, the angle $x/2$ is positive, meaning $\sin (x/2)$ is positive. We can simplify the expression to:

To use our standard limit, we multiply the denominator by $1/2$ and compensate by multiplying the whole expression by $1/2$. This gives us:

We know that for continuity, the RHL must also equal $\alpha$. So, we set the equation:

Solving for $\beta$, we get $\beta =2\sqrt{2}$. We have successfully determined both unknowns: $\alpha =2$ and $\beta =2\sqrt{2}$.

The final step is to calculate ${\alpha }^2+{\beta }^2$. Substituting our values:

We have bridged the gap, solved the mystery, and arrived at the elegant result of $12$.