Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

My teachers key says that it us -2. I was never taught how to evaluate limits that go to infinity or negative infinity so I need to be taught how to solve them.

My Teachers Key Says That It Us 2 I Was Never Taught How To Evaluate Limits That Go To Infinity Or Negative Infinity So I Need To Be Taught How To Solve Them class=

Sagot :

SOLUTION

The given limit is:

[tex]\lim_{x\to-\infty}\frac{\sqrt{4x^2+5}}{x-3}[/tex]

Divide by the numerator and denominator by x

[tex]\operatorname{\lim}_{x\to-\infty}\frac{\frac{\sqrt{4x^2+5}}{x}}{1-\frac{3}{x}}[/tex]

Upon simplifying this gives:

[tex]\begin{gathered} \frac{\lim_{x\to-\infty}\sqrt{\frac{4x^2}{x^2}-\frac{5}{x^2}}}{\lim_{n\to\infty}(1-\frac{3}{x})} \\ =\frac{\operatorname{\lim}_{x\to-\infty}\sqrt{4-\frac{5}{x^2}}}{\operatorname{\lim}_{n\to\infty}(1-\frac{3}{x})} \end{gathered}[/tex]

Taking the limit gives:

[tex]\begin{gathered} \frac{\sqrt{4+\frac{5}{-\infty}}}{1-\frac{3}{-\infty}} \\ \frac{\sqrt{4+0}}{1-0} \\ =\frac{\pm2}{1} \\ =\pm2 \end{gathered}[/tex]

Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.