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Calculating Limits Using Limit Laws and Rationalization | Sample from The Ultimate Crash Course Series

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Sample from The Ultimate Crash Course Series

This post is a sample lesson from The Ultimate Crash Course Series. Students who want the full bundle can get access to over 1,000 lessons, podcasts, and study resources through theSTEMmajor.com.

Calculating Limits Using Limit Laws and Rationalization

Learn how to evaluate a limit that first produces an indeterminate form by using limit laws and rationalization. This sample lesson from The Ultimate Crash Course Series is designed for students in precalculus, calculus, engineering, physics, and STEM programs who want stronger algebra and limit skills.

Problem

Find the limit, if it exists:

\displaystyle \lim_{u\to 2}\frac{\sqrt{4u+1}-3}{u-2}.

Step 1: Use Direct Substitution First

The first step in evaluating a limit is usually direct substitution. Replace u with 2:

\displaystyle \frac{\sqrt{4(2)+1}-3}{2-2}=\frac{\sqrt{9}-3}{0}=\frac{3-3}{0}=\frac{0}{0}.

Since direct substitution gives 0/0, the expression is in an indeterminate form. This means we must simplify the algebra before evaluating the limit.

Step 2: Rationalize the Numerator

Because the numerator contains a radical, multiply the fraction by the conjugate of the numerator:

\displaystyle \frac{\sqrt{4u+1}-3}{u-2}\cdot\frac{\sqrt{4u+1}+3}{\sqrt{4u+1}+3}.

This gives:

\displaystyle \frac{(\sqrt{4u+1}-3)(\sqrt{4u+1}+3)}{(u-2)(\sqrt{4u+1}+3)}.

Now simplify the numerator using the difference of squares formula:

\displaystyle (\sqrt{4u+1}-3)(\sqrt{4u+1}+3)=(4u+1)-9=4u-8.

So the expression becomes:

\displaystyle \frac{4u-8}{(u-2)(\sqrt{4u+1}+3)}.

Step 3: Factor and Cancel

Factor the numerator:

\displaystyle 4u-8=4(u-2).

Substitute that into the fraction:

\displaystyle \frac{4(u-2)}{(u-2)(\sqrt{4u+1}+3)}.

Cancel the common factor of u – 2:

\displaystyle \frac{4}{\sqrt{4u+1}+3}.

Step 4: Evaluate the Simplified Limit

Now substitute u = 2 into the simplified expression:

\displaystyle \lim_{u\to 2}\frac{4}{\sqrt{4u+1}+3}=\frac{4}{\sqrt{4(2)+1}+3}=\frac{4}{\sqrt{9}+3}=\frac{4}{3+3}=\frac{4}{6}=\frac{2}{3}.

Final Answer

\displaystyle \lim_{u\to 2}\frac{\sqrt{4u+1}-3}{u-2}=\frac{2}{3}.

Why This Limit Example Matters

This example shows an important calculus idea. When direct substitution produces 0/0, you should not stop. Instead, use algebraic tools such as rationalization, factoring, and cancellation to rewrite the expression into a form that can be evaluated.

Sample Lesson from The Ultimate Crash Course Series

This is just a sample from The Ultimate Crash Course Series. If you want the full bundle for math, calculus, physics, and engineering study, visit Payhip and explore the complete collection. Students can also access over 1,000 lessons and podcasts through theSTEMmajor.com.

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