Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to 0}\frac{(27+x{)}^{1/3}-3}{9-(27+x{)}^{2/3}}$ equals :

Welcome, aspiring engineer. Today, we are going to embark on a journey through the elegant world of limits. Limits are the foundation of calculus, the bridge between the finite and the infinite.

When you look at the problem

it might seem daunting. The fractional powers, the cube roots, and the square roots can feel overwhelming, but remember that every complex problem is just a collection of simple steps waiting to be unraveled.

Limits are not just about finding a value; they are about understanding behavior. We are asking, "What happens as we get closer and closer to a point, without actually touching it?"

This is the essence of calculus. It allows us to analyze functions at points where they might be undefined, such as our $\frac{0}{0}$ case. It is the mathematical way of peering into the unknown.

First, let's address the elephant in the room: the indeterminate form. In the world of limits, we never rush. We start by testing the waters by substituting $x=0$ into our expression.

The numerator becomes $(27+0{)}^{1/3}-3$, which is $3-3=0$. The denominator becomes $9-(27+0{)}^{2/3}$, which is $9-9=0$.

We have arrived at the classic $\frac{0}{0}$ form. This is not a dead end; it is a signal that there is a hidden factor waiting to be canceled.

How do we handle these fractional powers? This is where the art of substitution comes into play. We want to simplify the expression to make it look like something we recognize.

Let's define a new variable, $u=(27+x{)}^{1/3}$. By squaring both sides, we get:

Suddenly, the entire expression transforms. The numerator becomes $u-3$, and the denominator becomes $9-u^2$.

We must not forget the limit itself. When we change the variable from $x$ to $u$, we must also change the limit.

As $x\to 0$, $u=(27+x{)}^{1/3}$ approaches $(27+0{)}^{1/3}=3$. So, our limit is now:

Precision here is key. A small mistake in the limit bounds can lead to a completely wrong answer.

Now, let's look at the denominator: $9-u^2$. This is a classic difference of squares, $a^2-b^2=(a-b)(a+b)$.

Thus, $9-u^2=(3-u)(3+u)$. Our expression is now:

Notice the numerator is $u-3$ and the denominator has $(3-u)$. They differ by a factor of $-1$. We can rewrite $u-3$ as $-(3-u)$.

Always be mindful of the signs; it is the difference between a correct answer and a sign error.

Now, the magic happens. We cancel the $(3-u)$ term from the numerator and the denominator. We are left with:

The indeterminate form has vanished! We can now safely substitute $u=3$.

The final result is:

This journey shows us that even the most intimidating problems can be broken down into simple, manageable steps. Substitution is a powerful tool in your arsenal that allows you to see the structure beneath the complexity.

Keep practicing and questioning. Every time you solve a problem like this, you are training your mind to think like a mathematician. The final answer is $-\frac{1}{6}$.