Question

, find the indicated limit. In most cases, it will be wise to do some algebra first.. $$ \lim _{u \rightarrow 1} \frac{(3 u+4)(2 u-2)^{3}}{(u-1)^{2}} $$

Short answer

The limit is 0.

Step-by-step solution

Identify the Problem Type

The problem requires us to find the limit as \(u\) approaches 1 for the given expression. Seeing the direct substitution might lead to an indeterminate form \(\frac{0}{0}\), we must do some algebraic simplifications.

Simplify the Numerator

First, expand the binomial \((2u-2)^3\). This becomes:\[(2u-2)^3 = 8(u-1)^3\]Using this, the numerator becomes:\[(3u+4) \cdot 8(u-1)^3 \] which is equivalent to:\[8(3u+4)(u-1)^3.\]

Simplify the Denominator

The denominator is already a simple expression:\[(u-1)^2.\]

Cancelling Common Factors

Now, substitute the simplified numerator and denominator into the limit expression:\[\lim _{u \rightarrow 1} \frac{8(3u+4)(u-1)^3}{(u-1)^2}.\]Cancel \((u-1)^2\) in the denominator with part of the \((u-1)^3\) term in the numerator, which gives:\[\lim _{u \rightarrow 1} 8(3u+4)(u-1).\]

Evaluate the Limit

Now that we have:\[\lim _{u \rightarrow 1} 8(3u+4)(u-1),\]Directly substitute \(u = 1\):\[8(3(1)+4)(1-1) = 8 \times 7 \times 0 = 0.\]Thus, the limit evaluates to 0.

Key concepts

Algebraic Simplification

In calculus, algebraic simplification plays a crucial role, especially when finding limits. It involves manipulating expressions to make them simpler and easier to evaluate. By simplifying an expression, you might reveal cancellations or reductions that are not obvious at first glance. For instance, in the given exercise, the expression \[\frac{(3u+4)(2u-2)^3}{(u-1)^2}\] at first appears complex. By manipulating it through algebraic simplification, one can streamline the expression into a more manageable form. Specifically, simplifying the expression (2u-2)^3 to 8(u-1)^3 facilitates identifying common terms that will cancel out later when evaluating the limit.

Indeterminate Forms

Indeterminate forms often arise when directly substituting a value into a limit expression gives a 0/0 form, or other undefined results like \(\infty/\infty\). Recognizing these forms is essential in calculus as they necessitate further algebraic manipulation or the use of limit properties to resolve. In the problem, substituting \(u = 1\) initially results in 0/0. This signals an indeterminate form, prompting algebraic simplification to proceed further. Typically, eliminating indeterminate forms involves factoring, expanding, or rationalizing the expression to reveal common terms that cancel out or otherwise simplify the expression. This process transforms the indeterminate form into a determinate one, making the limit evaluable.

Binomial Expansion

Binomial expansion is a powerful algebraic tool that allows us to express powers of binomials as a sum of terms. This is crucial in calculus, especially when dealing with polynomials or expressions of the form \((a + b)^n\). The binomial theorem states: \[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\]In the limit problem at hand, the term \((2u-2)^3\) can be expanded applying the formula. For simplicity, it was recognized as \(8(u-1)^3\). This expansion makes later steps, such as canceling terms in indeterminate forms, much more straightforward and improves our ability to handle complex algebraic expressions efficiently.

Limit Evaluation Steps

Evaluating limits involves several crucial steps, especially when dealing with indeterminate forms. Knowing these steps is key to solving limit problems efficiently.

• Identify Indeterminate Forms: Always begin by direct substitution to check if the outcome is an indeterminate form, like 0/0.
• Simplify the Expression: Use techniques such as factoring or expanding to rewrite the expression. This often uncovers cancellations.
• Cancel Common Factors: Identify and cancel terms that exist in both the numerator and denominator if applicable.
• Re-evaluate the Limit: Once simplified, substitute the limiting value again. This should yield a valid numerical answer.

In our exercise, canceling \((u-1)^2\) in the denominator with a part of \((u-1)^3\) in the numerator simplifies the expression sufficiently to allow direct substitution, leading to the final result.