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Calculus Name ID: 1 Assignment - 2016 Kuta Software LL,C. All rights reserve...

Dec 19, 2023

\lim _{x \rightarrow \infty} \frac{x+2}{\sqrt{2 x^{2}+2}}

\lim _{x \rightarrow \infty} \frac{\sqrt{2 x^{2}+1}}{2 x+1}

\lim _{x \rightarrow \infty} \frac{4 x-1}{\sqrt{x^{2}+3}}

\lim _{x \rightarrow+\infty} \frac{\sqrt{x^{2}+2}}{x+1}

\lim _{x \rightarrow \infty} \sqrt[3]{\frac{4 x^{2}+3}{x^{2}+1}}

\lim _{x \rightarrow-} \sqrt[3]{\frac{3 x^{2}+3}{x^{2}+4}}

\lim _{x \rightarrow-\infty} \sqrt{\frac{2 x^{2}+2}{2 x^{2}+1}}

\lim _{x \rightarrow-\infty} \frac{\sqrt{4 x^{2}+2}}{4 x}

\lim _{x \rightarrow-\infty} \frac{2 x-1}{\sqrt{3 x^{2}+2}}

10)

\lim _{x \rightarrow \infty} \sqrt{\frac{3 x^{2}+5}{2 x^{2}+3}}

11)

\lim _{x \rightarrow \infty} \frac{3 x-1}{\sqrt{3 x^{2}+3}}

12)

\lim _{x \rightarrow-\infty} \frac{3 x+2}{\sqrt{4 x^{2}+3}}

Solution by Steps

step 1

To evaluate the limit

\lim _{x \rightarrow \infty} \frac{x+2}{\sqrt{2 x^{2}+2}}

, we divide the numerator and the denominator by the highest power of

x

in the denominator, which is

x

step 2

Simplifying the expression:

\lim _{x \rightarrow \infty} \frac{\frac{x}{x}+\frac{2}{x}}{\sqrt{\frac{2 x^{2}}{x^{2}}+\frac{2}{x^{2}}}} = \lim _{x \rightarrow \infty} \frac{1+\frac{2}{x}}{\sqrt{2+\frac{2}{x^{2}}}}

step 3

x

approaches infinity, the terms

\frac{2}{x}

and

\frac{2}{x^{2}}

approach 0:

\lim _{x \rightarrow \infty} \frac{1+0}{\sqrt{2+0}} = \frac{1}{\sqrt{2}}

Answer

\frac{1}{\sqrt{2}}

Key Concept

Dividing by the highest power of

x

in the denominator to evaluate limits at infinity

Explanation

By dividing by the highest power of

x

in the denominator, we simplify the expression to find the limit as

x

approaches infinity. Terms with

x

in the denominator approach 0, leaving the constant terms.

Repeat this format for each of the remaining questions.