91Ó°ÊÓ

Indeterminate forms in calculus are expressions that do not have a clear value without further analysis. One of the most common indeterminate forms is \(\frac{0}{0}\). When dealing with limits, if both the numerator and the denominator approach zero as the variable approaches a certain value, like in our given problem, we end up with an indeterminate form.

Knowing that an expression is in an indeterminate form is essential. It signals that we must simplify or manipulate the expression to find a definite limit. Without recognizing these forms, computing limits accurately can become almost impossible.

Other typical indeterminate forms include \(\frac{\text{∞}}{\text{∞}}, \text{∞} - \text{∞}, 0 \times \text{∞}, 1^\text{∞}, 0^0, \text{ and } \text{∞}^0\).

Factoring expressions is a key technique in simplifying functions, especially when computing limits. In the given exercise, the numerator \(2x - 10\) and the denominator \(x^2 - 25\) need to be factored to facilitate simplification.

Factoring the numerator: \(2x - 10\) can be written as \(2(x - 5)\).

Factoring the denominator: \(x^2 - 25\) is a difference of squares and can be expressed as \((x - 5)(x + 5)\).

This step is critical because it reveals common factors in the numerator and denominator that can be canceled out. Only after factoring correctly can we move on to simplifying the given expression to compute the limit effectively.

Simplifying fractions involves canceling out common factors in the numerator and the denominator. In our problem, once we factor both the numerator and the denominator, we get: \(\frac{2(x - 5)}{(x - 5)(x + 5)}\).

Here, \(x - 5\) is a common factor in both the numerator and the denominator. By canceling \(x - 5\), we simplify the fraction to: \(\frac{2}{x + 5}\).

Simplifying fractions is crucial because it removes the indeterminate form and makes it possible to directly substitute the value of the variable to find the limit. Always be cautious while canceling factors. Ensure that the factor you're canceling is common and correctly factored out.

After simplifying the fraction, the final step is to compute the limit by substituting the variable with the value it approaches. In our problem, after simplification, the expression becomes \(\frac{2}{x + 5}\).

Now, we'll substitute \(x = 5\) into the simplified expression: \(\frac{2}{5 + 5} = \frac{2}{10} = \frac{1}{5}\).

Computing the limit involves evaluating the simplified expression at the given point. This concludes the limit problem, giving us a definite value. By simplifying the expression first, it becomes straightforward to compute the limit and avoid indeterminate forms throughout the process.