Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f(x)=\sqrt{\frac{x-\sin x}{x+{\cos }^{2}x}}$, then ${\lim }_{x\to \infty }f(x)$ is

Imagine you are standing on the $x$-axis, and you start walking towards the right, moving further and further away, past every number you can count, towards infinity. You are looking at the function:

At first glance, it looks intimidating—a square root, algebraic terms, and trigonometric functions all mixed together. But do not let the complexity fool you. In the world of calculus, infinity is not a destination to fear; it is a simplification tool.

When we evaluate limits as $x\to \infty$, we are essentially asking: "What is the dominant behavior of this function when $x$ becomes massive?"

The trigonometric terms $\sin x$ and ${\cos }^{2}x$ are what we call "bounded functions." They are trapped, oscillating forever between $-1$ and $1$ (or $0$ and $1$). They are like tiny ripples in an ocean, while $x$ is the ocean itself.

As $x$ grows, these ripples become insignificant. To see this mathematically, we perform a bit of algebraic surgery. We divide both the numerator and the denominator inside the square root by the highest power of $x$, which is $x^1$:

Now, let us simplify this. Distributing the division, we get:

This is where the magic happens. We know that $-1\le \sin x\le 1$. If we divide this inequality by $x$, we get:

As $x\to \infty$, both $-\frac{1}{x}$ and $\frac{1}{x}$ approach $0$. By the Squeeze Theorem, ${\lim }_{x\to \infty }\frac{\sin x}{x}=0$.

The same logic applies to $\frac{{\cos }^{2}x}{x}$. Since $0\le {\cos }^{2}x\le 1$, we have $0\le \frac{{\cos }^{2}x}{x}\le \frac{1}{x}$. Again, as $x\to \infty$, this term is squeezed to $0$.

With these terms vanishing into the void of zero, our expression becomes incredibly simple:

This leaves us with $\sqrt{\frac{1}{1}}$, which is simply $1$.

We have successfully navigated the complexity and found that as $x$ approaches infinity, our function settles down to the value of $1$. Geometrically, this means the graph of our function has a horizontal asymptote at $y=1$.