If \(\mathrm{f}(\mathrm{x})=\left[\left\\{2+(256+7 \mathrm{x})^{(1 / 8)}\right\\} /\left\\{(5 \mathrm{x}+32)^{(1 / 5)}-2\right\\}\right](\mathrm{x} \neq 0)\), then for \(\mathrm{f}\) to be continuous everywhere \(\mathrm{f}(0)\) is equal to (a) \((2 / 7)\) (b) \(-(7 / 32)\) (c) \((7 / 64)\) (d) \(-(7 / 64)\)

Understanding the concept of limits is a foundational element in calculus, which deals with the behavior of functions as inputs approach a certain point. A limit attempts to describe mathematically what the value of a function approaches as the input gets close to a specific value. The formal definition is rather complex, but the general idea is to see what value the function tends toward as the variable approaches a particular point.

When calculating limits, like the one in our exercise \( \lim_{x\to 0} f(x) \), our goal is to find out what value the function \( f(x) \) is getting closer to as \( x \) gets very close to 0 but never actually reaches it. In some cases, the function approaches a specific number, which is the limit. In other exercises, the function might not approach a single value, or it might approach a different value depending on the direction from which \( x \) is coming.

In the step-by-step solution provided, we saw that after simplifying the function and substituting \( x \) with 0, we found that \( \lim_{x\to 0} f(x) \) equals 2. This step is crucial as it tests whether the function has a definite trend as \( x \) approaches 0.

Function continuity is a property that describes whether a function is smooth and unbroken over its domain. In other words, you can draw the graph of a continuous function without lifting your pencil from the paper. For a function to be continuous at a point, three conditions must be met:

• The function must be defined at the point.
• The limit of the function as it approaches the point from either direction must exist.
• The limit of the function at that point must equal the actual value of the function at that point.

In our exercise, we are checking for continuity at \( x = 0 \). The function \( f(x) \) is continuous everywhere except potentially at \( x = 0 \) because it's not defined there. To make \( f(x) \) continuous at \( x = 0 \), we need to assign a value of \( f(0) \) such that it equals the limit of \( f(x) \) as \( x \) approaches 0. From our solution, this value should be 2, but that isn't an option in the exercise, indicating a possible error in the question.

Limit properties are rules that allow us to manipulate and calculate limits in a more straightforward manner. They make it possible to break down complicated limits into simpler parts that are easier to manage. Some of the fundamental properties used in calculations involve sums, differences, products, quotients, and roots of limits.

In the solution provided, several limit properties come into play. For example, the property that \( \lim_{x\to a} [f(x) + g(x)] = \lim_{x\to a} f(x) + \lim_{x\to a} g(x) \) allows us to break apart the limit of a sum of two functions into the sum of their limits. Additionally, we use the property that the limit of a constant is the constant itself and that the limits of polynomial functions and roots are straightforward to compute.

These properties streamlined our process in finding \( \lim_{x\to 0} f(x) \) by allowing us to simplify the function and cancel out common terms, leading to the correct calculation of the limit as \( x \) approaches 0. It's important to become familiar with these properties, as they are essential tools in solving limit problems across calculus.