Solving Limits with Indeterminate Forms: A Step-by-Step Guide

When dealing with limits, sometimes you encounter indeterminate forms such as (frac{0}{0}) or (frac{infty}{infty}). These forms arise when both the numerator and the denominator approach zero, leading to an undefined result. This article will guide you through the process of solving such limits.

Introduction to Indeterminate Forms

In calculus, indeterminate forms refer to situations where the standard limit rules are insufficient to find the limit of a function. The most common indeterminate forms encountered are:

(frac{0}{0}) (frac{infty}{infty}) 00, 1(infty), and (infty^0) 0(cdot)(infty) (infty - infty) (frac{infty}{infty})

Understanding these forms and how to resolve them is crucial in calculus. This article will focus on resolving the (frac{0}{0}) form using algebraic manipulation and factoring techniques.

Example: Solving the Limit

Consider the limit problem:

    lim_{x to 4} frac{2x^3-128}{sqrt{x}-2}

This limit initially evaluates to the indeterminate form (frac{0}{0}):

Step 1: Initial Evaluation

Let's substitute x 4 directly into the expression:

Numerator: (2(4)^3 - 128 128 - 128 0) Denominator: (sqrt{4} - 2 2 - 2 0)

Since both the numerator and denominator evaluate to zero, we have the indeterminate form (frac{0}{0}).

Step 2: Factor the Numerator

The expression in the numerator can be factored using the difference of cubes formula:

[2x^3 - 128 2(x^3 - 64) 2(x - 4)(x^2 4x 16)]

Therefore, the limit becomes:

    lim_{x to 4} frac{2(x - 4)(x^2   4x   16)}{sqrt{x} - 2}

Step 3: Rewrite the Denominator

The denominator can be rewritten using the conjugate:

[sqrt{x} - 2 frac{(sqrt{x} - 2)(sqrt{x} 2)}{(sqrt{x} 2)} frac{x - 4}{sqrt{x} 2}]

Substituting this back into the limit:

    lim_{x to 4} frac{2(x - 4)(x^2   4x   16)}{frac{x - 4}{sqrt{x}   2}}

Simplifying, we get:

    lim_{x to 4} 2(x^2   4x   16)(sqrt{x}   2)

Step 4: Cancel Common Terms

We can now cancel the common factor of (x - 4) in the numerator and denominator:

    lim_{x to 4} 2(x^2   4x   16)(sqrt{x}   2)

Step 5: Evaluate the Limit

Substituting x 4:

(x^2 4x 16) at x 4) gives 4^2 4(4) 16 16 16 16 48) (sqrt{x} 2) at x 4) gives (sqrt{4} 2 2 2 4)

Therefore, the limit evaluates to:

[2(48)(4) 384]

Thus, the limit is:

[boxed{384}]

Conclusion

By carefully factoring the numerator and rewriting the denominator using the conjugate, we were able to resolve the indeterminate form and find the limit. This method is particularly useful in calculus and forms the basis for solving more complex limit problems.