Question
3. [-/0.83 Points] DETAILS MY NOTES SCALCET9 2.XP.6.008. Find the limit. (If the limit is infinite, enter ' \( \infty \) ' or '- \( \infty \) ', as appropriate. If the limit does not otherwise exist, enter DNE.) \( \lim _{x \rightarrow \infty} \frac{x^{2}}{\sqrt{x^{4}+9}} \)
Ask by Johnston Carrillo. in the United States
Feb 26,2025
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Tutor-Verified Answer
Answer
The limit is 1.
Solution
Calculate the limit \( \lim_{x\rightarrow \infty} \frac{x^{2}}{\sqrt{x^{4}+9}} \).
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
\(\lim _{x\rightarrow +\infty}\left(\frac{x^{2}}{\sqrt{x^{4}+9}}\right)\)
- step1: Rewrite the expression:
\(\lim _{x\rightarrow +\infty}\left(\frac{x^{2}}{\sqrt{1+\frac{9}{x^{4}}}\times x^{2}}\right)\)
- step2: Reduce the fraction:
\(\lim _{x\rightarrow +\infty}\left(\frac{1}{\sqrt{1+\frac{9}{x^{4}}}}\right)\)
- step3: Rewrite the expression:
\(\frac{\lim _{x\rightarrow +\infty}\left(1\right)}{\lim _{x\rightarrow +\infty}\left(\sqrt{1+\frac{9}{x^{4}}}\right)}\)
- step4: Calculate:
\(\frac{1}{\lim _{x\rightarrow +\infty}\left(\sqrt{1+\frac{9}{x^{4}}}\right)}\)
- step5: Calculate:
\(\frac{1}{1}\)
- step6: Divide the terms:
\(1\)
The limit of \( \frac{x^{2}}{\sqrt{x^{4}+9}} \) as \( x \) approaches infinity is 1.
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Beyond the Answer
To solve the limit \( \lim _{x \rightarrow \infty} \frac{x^{2}}{\sqrt{x^{4}+9}} \), we start by analyzing the expression. Notice that as \( x \) approaches infinity, the term \( x^{4} \) inside the square root dominates the 9. Thus, we can simplify the limit: \[ \lim _{x \rightarrow \infty} \frac{x^{2}}{\sqrt{x^{4}+9}} \approx \lim _{x \rightarrow \infty} \frac{x^{2}}{\sqrt{x^{4}}} = \lim _{x \rightarrow \infty} \frac{x^{2}}{x^{2}} = 1. \] So the limit is equal to 1. The answer is \( 1 \).