Animated Solution for Mathematics - Limits, Continuity and Differentiability: If for $p\ne q\ne 0$, then function $f(x)=\frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+qx}-9}$ is continuous at $x=0$, then:

Welcome, future engineer! Today, we are going to dismantle a problem that looks intimidating at first glance but hides a beautiful, elegant structure. When you see a function defined with seventh and third roots, it is natural to feel a moment of hesitation.

But remember, in the world of JEE Advanced, complexity is often just a mask for simplicity. Let us peel back that mask together.

We are given a function:

We are told it is continuous at $x=0$. Continuity implies that the bridge is unbroken; the value of the function at $x=0$ must be equal to the limit of the function as $x$ approaches $0$.

Mathematically, this is expressed as:

Let us test the denominator first. As $x\to 0$, the term $qx$ vanishes, leaving us with $\sqrt[3]{729}-9$. Since $9^3=729$, this results in $9-9=0$.

We have a zero in the denominator! If the numerator were any non-zero number, the limit would explode to infinity, and continuity would be impossible.

Therefore, for the limit to exist, the numerator must also approach zero. This is our first key condition:

With the condition established, we substitute $x=0$ into the numerator:

Raising both sides to the power of $7$, we get $729p=3^7$. Now, note that $729=3^6$.

Our equation becomes $3^6\cdot p=3^7$. Dividing both sides by $3^6$, we find the elegant result:

Now that we have $p=3$, our function becomes:

To evaluate this limit, we use the binomial approximation $(1+z{)}^n\approx 1+nz$ for small $z$. We must force the terms inside the roots to look like $(1+z)$.

Factor out $729$ from inside the roots. For the numerator:

Since $3\cdot 729=3\cdot 3^6=3^7$, the seventh root of $3^7$ is $3$. Thus, the numerator becomes $3(1+\frac{x}{729}{)}^{\frac{1}{7}}-3$.

Similarly, the denominator becomes $9(1+\frac{qx}{729}{)}^{\frac{1}{3}}-9$.

Now, apply the binomial approximation. The numerator becomes:

The denominator becomes:

Look at the beauty of the final limit:

The $x$ terms, the $3$s, and the $729$s all cancel out. We are left with:

We have $f(0)=\frac{1}{7q}$, which implies $7qf(0)=1$. Since $p=3$, we know $p^2=9$.

Multiplying our equation by $9$, we get $63qf(0)=9$. Substituting $p^2$ for $9$, we arrive at the final relation:

You have successfully navigated the complexity and found the truth hidden within the algebra. Keep this confidence; you are ready for the next challenge!