Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of ${\lim }_{x\to 0}((\sin x{)}^{1/x}+(1/x{)}^{\sin x})$, where $x>0$ is

Taming the Monster Limit

Welcome, future engineer. Today, we are going to dismantle a problem that looks like a nightmare but is actually a beautiful exercise in logical decomposition.

We are tasked with finding the limit:

When you see a complex expression like this, the first instinct is often panic. But remember, in JEE Advanced, the most complex problems are often just several simple problems wearing a trench coat. Let's strip away the disguise.

Divide and Conquer

We have a sum of two distinct terms. The limit of a sum is the sum of the limits, provided those limits exist.

So, let us define:

Our goal is simply to find $L_1$ and $L_2$ and add them together. By splitting the problem, we have already reduced our cognitive load by half.

The $0^{\infty }$ Trap

Let's look at $L_1={\lim }_{x\to 0^{+}}(\sin x{)}^{1/x}$. As $x$ approaches zero from the positive side, $\sin x$ approaches zero. Simultaneously, the exponent $1/x$ approaches infinity.

We are looking at the form $0^{\infty }$. Now, pause. Is this indeterminate? No.

Think about it physically. You are taking a tiny fraction—say, $0.0001$—and multiplying it by itself an infinite number of times. It doesn't fight back; it simply vanishes into zero. Thus, $L_1=0$.

The Logarithmic Transformation

Now for the real challenge: $L_2={\lim }_{x\to 0^{+}}(1/x{)}^{\sin x}$. As $x\to 0^{+}$, the base $1/x$ goes to infinity, and the exponent $\sin x$ goes to zero.

This is the classic ${\infty }^0$ indeterminate form. We need a tool to bring that exponent down to earth. We use the identity $a^b=e^{b\ln a}$.

Applying this, we get:

Because the exponential function is continuous, we can move the limit inside the exponent:

The L'Hôpital Dance

Let's focus on the exponent limit, $k={\lim }_{x\to 0^{+}}\sin x\ln (1/x)$. Using the property $\ln (1/x)=-\ln x$, we get $k=-{\lim }_{x\to 0^{+}}\sin x\ln x$.

This is a $0\times \infty$ form. To use L'Hôpital's Rule, we need a fraction. Let's rewrite it as:

Now we have the $\infty /\infty$ form. Differentiating the numerator gives $1/x$, and the denominator gives $-\csc x\cot x$. The negative signs cancel, and we are left with:

The Grand Finale

Let's simplify this expression. Converting to sines and cosines, we get:

We can split this into:

As $x\to 0$, the first part is a standard limit equal to $1$, and the second part is $\tan x$, which approaches $0$. So, $k=1\cdot 0=0$.

Finally, substituting back into $L_2=e^k$, we get $L_2=e^0=1$. Adding our results, $L=0+1=1$. We have tamed the monster!