Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\{\begin{array}{ll}\frac{1-\cos 4x}{x^2},& x<0\\ a,& x=0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4},& x>0\end{array}$. Determine the value of $a$, if possible, so that the function is continuous at $x=0$.

Mastering Continuity

Imagine you are an engineer tasked with building a bridge. You have two separate roads—one coming from the left and one from the right—and you need to connect them at a single point, $x=0$.

If the roads don't meet at the exact same height, you have a disaster: a jump, a break, a discontinuity. In calculus, we call this the condition of continuity.

For our function $f(x)$ to be continuous at $x=0$, the path from the left (the Left Hand Limit) and the path from the right (the Right Hand Limit) must converge to the exact same destination, and the value of the function at that point, $f(0)=a$, must be that very same destination. Let's embark on this journey to find $a$.

The Left-Hand Limit - Unmasking the Trigonometric Identity

We begin by approaching zero from the negative side. Our function is defined as:

If we try to be impatient and plug in $x=0$, we get $1-\cos (0)=0$ in the numerator and $0^2=0$ in the denominator. A $0/0$ indeterminate form! This is the universe telling us that the function is hiding its true value.

Whenever you see $1-\cos \theta$, your mathematical intuition should immediately scream: "Half-angle identity!" We know that $1-\cos \theta =2{\sin }^2(\frac{\theta }{2})$.

In our case, $\theta =4x$, so $\frac{\theta }{2}=2x$. Our expression transforms into:

Now, we want to use the golden rule of limits: ${\lim }_{\theta \to 0}\frac{\sin \theta }{\theta }=1$. To make our expression look like this, we need the denominator to be $(2x{)}^2$, which is $4x^2$.

We multiply and divide by $4$ to get:

As $x\to 0$, the term $\frac{\sin (2x)}{2x}$ approaches $1$. Thus, our Left Hand Limit is $8\cdot (1{)}^2=8$. The road from the left is heading straight for $y=8$.

The Right-Hand Limit - The Rationalization Rescue

Now, we turn to the right side, where:

Again, direct substitution yields $0/0$. This time, the culprit is the square root in the denominator. The most powerful tool in our arsenal here is rationalization.

We multiply the numerator and the denominator by the conjugate: $\sqrt{16+\sqrt{x}}+4$. This creates a difference of squares in the denominator:

Look at the beauty of this collapse! The $\sqrt{x}$ in the denominator now perfectly cancels with the $\sqrt{x}$ in the numerator. We are left with:

Substituting $x=0$ is now trivial: $\sqrt{16}+4=4+4=8$. The road from the right is also heading straight for $y=8$.

Assembling the Bridge

We have done the hard work. The left road approaches $8$, and the right road approaches $8$.

For the bridge to be continuous, the value of the function at the junction, $f(0)=a$, must be exactly $8$. If $a$ were anything else, there would be a hole or a jump, and our bridge would fail.

By setting $a=8$, we perfectly fill the gap. You have successfully navigated the indeterminate forms and ensured the continuity of the function. This is the elegance of calculus—taking complex, seemingly broken expressions and finding the singular value that brings them into perfect harmony.