Animated Solution for Mathematics - Limits, Continuity and Differentiability: If ${\lim }_{x\to 1}\frac{(5x+1{)}^{1/3}-(x+5{)}^{1/3}}{(2x+3{)}^{1/2}-(x+4{)}^{1/2}}=\frac{m\sqrt{5}}{n(2n{)}^{2/3}}$, where $\gcd (m,n)=1$, then $8m+12n$ is equal to ______

Welcome, future engineer! Today, we are going to conquer a limit problem that looks intimidating but is actually a beautiful dance of calculus.

We are evaluating the following expression:

When you first look at this, your instinct might be to panic at the fractional powers. But take a breath. In the world of JEE Advanced, the first step is always the same: test the waters.

By substituting $x=1$, we find that the numerator becomes $6^{1/3}-6^{1/3}=0$, and the denominator becomes $5^{1/2}-5^{1/2}=0$. We have hit the $\frac{0}{0}$ indeterminate form.

This is not a wall; it is a gate. It tells us that we have a hidden factor of $(x-1)$ lurking in both the numerator and the denominator, and we have the perfect key to unlock it: L'Hopital's Rule.

Applying L'Hopital's Rule

L'Hopital's Rule is your most reliable sword in the calculus arena. It states that for a $\frac{0}{0}$ form, the limit of the ratio is simply the limit of the ratio of the derivatives.

We must differentiate the numerator $N(x)=(5x+1{)}^{1/3}-(x+5{)}^{1/3}$ and the denominator $D(x)=(2x+3{)}^{1/2}-(x+4{)}^{1/2}$ with respect to $x$. This is where the Chain Rule becomes our best friend.

Applying the rule where the derivative of $(ax+b{)}^n$ is $n(ax+b{)}^{n-1}\cdot a$, we obtain:

Similarly, for the denominator, we get:

Simplifying the Chaos

Now that we have our derivatives, we evaluate them at $x=1$. For the numerator:

For the denominator:

Now, we assemble the pieces. The limit $L$ is defined as:

With a bit of algebraic rearrangement, we flip the denominator and move the negative powers to their rightful places:

Matching the Form

We are almost at the finish line. The problem asks us to match our result to the form $\frac{m\sqrt{5}}{n(2n{)}^{2/3}}$. We have $\frac{8\sqrt{5}}{3\cdot 6^{2/3}}$.

If we look at the denominator, we see $3\cdot 6^{2/3}$. We need it to look like $n(2n{)}^{2/3}$. If we set $n=3$, then $2n=6$.

The expression becomes $3(6{)}^{2/3}$, which is a perfect match! Thus, $m=8$ and $n=3$. The condition $\gcd (8,3)=1$ is satisfied.

Finally, we calculate the requested value:

You have navigated the complexity, applied the rules with precision, and arrived at the truth. The final answer is 100.