Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f(9)=9,f^{\prime }(9)=4$, then ${\lim }_{x\to 9}\frac{\sqrt{f(x)}-3}{\sqrt{x}-3}$ equals .........

Welcome, future engineers! Today, we are going to peel back the layers of a classic limit problem. We are given $f(9)=9$ and $f^{\prime }(9)=4$, and we need to evaluate the following limit:

The first rule of limits is always direct substitution. When we plug in $x=9$, we get:

This is the infamous indeterminate form. It is not a wall; it is a door. It tells us that there is a hidden factor of $(x-9)$ lurking in both the numerator and the denominator, waiting to be cancelled out.

How do we find that hidden factor? We see square roots, and our instinct should immediately jump to rationalization to clear the roots and reveal the underlying algebra.

We multiply the numerator and the denominator by their respective conjugates:

Rearranging the terms, we obtain:

Now, we apply the difference of squares identity, $(a-b)(a+b)=a^2-b^2$. The numerator becomes $f(x)-9$, and the denominator becomes $x-9$. Our limit now looks like this:

This is where the magic happens. Look at the first part of our product: ${\lim }_{x\to 9}\frac{f(x)-9}{x-9}$.

Since we know $f(9)=9$, we can rewrite this as:

This is the exact definition of the derivative $f^{\prime }(9)$. We are given that $f^{\prime }(9)=4$, so the first part of our limit is simply $4$.

Now, for the second part, ${\lim }_{x\to 9}\frac{\sqrt{x}+3}{\sqrt{f(x)}+3}$, we can safely substitute $x=9$ because the denominator is no longer zero:

Finally, we multiply our two results: $4\cdot 1=4$. The complexity vanishes, leaving behind a clean, elegant integer.

The final answer is 4. Remember, in JEE Advanced, the math is rarely just about calculation; it is about recognizing the fundamental definitions hidden in plain sight.