Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to 0^{+}}\frac{\tan (5(x{)}^{1/3}){\log }_e(1+3x^2)}{({\tan }^{-1}3\sqrt{x}{)}^2(e^{5(x{)}^{4/3}}-1)}$ is equal to

A Journey of Simplification

Have you ever looked at a limit problem and felt like you were staring at a tangled knot of math? You see fractional powers, trigonometric functions, logarithms, and exponentials all fighting for space. It is easy to feel overwhelmed.

But today, we are going to learn how to untangle that knot. We are going to take this expression:

We will break it down until it is so simple, you will wonder why it ever looked scary.

Before we touch the algebra, let us talk about the philosophy of limits. When we see a limit as $x\to 0$, we are looking for the behavior of a function in the immediate neighborhood of zero.

In this tiny, microscopic region, complex functions like $\tan \theta$ or ${\log }_e(1+u)$ behave almost exactly like simple linear functions. This is the secret weapon of the JEE topper.

We are not guessing; we are using the first term of the Taylor series expansion. We are effectively saying: "In the limit, this complex function is just a line."

Let us look at our numerator: $\tan (5x^{\frac{1}{3}}){\log }_e(1+3x^2)$. As $x\to 0^{+}$, the argument $5x^{\frac{1}{3}}$ approaches zero.

Our rule $\tan \theta \approx \theta$ tells us that $\tan (5x^{\frac{1}{3}})\approx 5x^{\frac{1}{3}}$. Now, look at the log term: ${\log }_e(1+3x^2)$.

As $x\to 0^{+}$, the argument $3x^2$ also approaches zero. Our rule ${\log }_e(1+u)\approx u$ tells us that ${\log }_e(1+3x^2)\approx 3x^2$.

Just like that, the numerator has transformed from a trigonometric-logarithmic mess into a clean algebraic product: $(5x^{\frac{1}{3}})\cdot (3x^2)$.

Now, let us turn to the denominator: $({\tan }^{-1}3\sqrt{x}{)}^2(e^{5x^{\frac{4}{3}}}-1)$. This is where many students stumble. Let us take it slow.

First, the inverse tangent: ${\tan }^{-1}3\sqrt{x}$. As $x\to 0^{+}$, $3\sqrt{x}\to 0$. So, ${\tan }^{-1}3\sqrt{x}\approx 3\sqrt{x}$.

But wait! The entire term is squared. We must square the result: $(3\sqrt{x}{)}^2=9x$. Do not let that square catch you off guard!

Next, the exponential term: $e^{5x^{\frac{4}{3}}}-1$. Since the exponent $5x^{\frac{4}{3}}$ approaches zero, we use the rule $e^w-1\approx w$. So, this term becomes $5x^{\frac{4}{3}}$.

The denominator is now $9x\cdot 5x^{\frac{4}{3}}$.

We have successfully stripped away the complexity. Our limit is now:

Let us clean up the numerator: $5\cdot 3=15$, and $x^{\frac{1}{3}}\cdot x^2=x^{\frac{1}{3}+\frac{6}{3}}=x^{\frac{7}{3}}$. So, the numerator is $15x^{\frac{7}{3}}$.

Now for the denominator: $9\cdot 5=45$, and $x^1\cdot x^{\frac{4}{3}}=x^{\frac{3}{3}+\frac{4}{3}}=x^{\frac{7}{3}}$. The denominator is $45x^{\frac{7}{3}}$.

Look at what we have:

The $x^{\frac{7}{3}}$ terms cancel out perfectly! We are left with $\frac{15}{45}$, which simplifies to $\frac{1}{3}$.

What started as a terrifying expression has been reduced to a simple fraction. This is the beauty of mathematics—no matter how complex the problem looks, there is always a path to simplicity if you know which tools to use.