Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f(x)=\{\begin{array}{ll}\sin x,& x\ne n\pi \\ 2,& \text{otherwise}\end{array}$ and $g(x)=\{\begin{array}{ll}{x}^2+1,& x\ne 0,2\\ 4,& x=0\\ 5,& x=2\end{array}$, then ${\lim }_{x\to 0}g[f(x)]$ is .........

A Journey, Not a Destination

Welcome, fellow traveler on the road to JEE Advanced mastery. Today, we are not just solving a problem; we are dissecting the very soul of calculus.

The problem before us, ${\lim }_{x\to 0}g[f(x)]$, is a classic. It is designed to test whether you truly understand the definition of a limit or if you are merely following rote algorithms. Let us peel back the layers of this composite function.

The Inner Function and the Neighborhood

Imagine you are standing on the $x$-axis, walking toward the origin ($x=0$). You are getting closer and closer, but you never actually step on the zero. This is the essence of the limit.

When we look at our inner function, $f(x)$, we see a piecewise definition:

As we approach $x=0$, we are in a tiny neighborhood where $x$ is definitely not an integer multiple of $\pi$. Therefore, the condition $x eq n\pi$ is satisfied. We can confidently say that in this neighborhood, $f(x)=\sin x$.

Notice the trap: at exactly $x=0$, $f(0)=2$. But because we are taking a limit, we ignore this isolated point. It is a distraction, a siren song meant to lure you into the wrong branch. We stay focused on the path, not the destination.

The Composite Variable $u$

Let us simplify our mental landscape. Let $u=f(x)$. As $x\to 0$, we know that $f(x)=\sin x$.

We know from our trigonometric foundations that:

So, as $x$ approaches $0$, our new variable $u$ approaches $0$. But here is the critical, JEE-level insight: $u$ is not exactly $0$.

Because $x$ is not exactly $0$, $\sin x$ is not exactly $0$. It is a value incredibly close to $0$, but non-zero. This distinction is the difference between a correct answer and a catastrophic error. We are now looking for ${\lim }_{u\to 0}g(u)$.

The Outer Function and the Final Trap

Now we turn our attention to the outer function, $g(u)$. We are given $g(u)=u^2+1$ for $u eq 0, 2$, and $g(0)=4$. We are evaluating the limit as $u\to 0$.

Since $u$ is approaching $0$ but is not equal to $0$, we must use the branch $g(u)=u^2+1$. If we had mistakenly used $g(0)=4$, we would have fallen into the trap.

The limit does not care that $g(0)=4$; it only cares about the values of $g(u)$ as $u$ gets closer and closer to $0$. The function $g(u)$ is effectively $u^2+1$ in the neighborhood of $0$.

The Elegant Conclusion

We have arrived at the final step. We need to evaluate ${\lim }_{u\to 0}(u^2+1)$.

Since the polynomial $u^2+1$ is continuous at $u=0$, we can now perform direct substitution. Substituting $u=0$ into $u^2+1$, we get:

Alternatively, if we substitute back $u=\sin x$, we are evaluating ${\lim }_{x\to 0}({\sin }^{2}x+1)$. As $x\to 0$, $\sin x\to 0$, so ${\sin }^{2}x\to 0$, and the expression approaches $0+1=1$.

The result is 1. It is elegant, it is precise, and it is the result of respecting the fundamental definition of a limit. You have navigated the traps, ignored the isolated points, and arrived at the truth. Keep this level of precision in your toolkit, and no JEE problem will ever be able to stand in your way.