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Lim=03x3+2x2+x+1x3+2x+5

Sagot :

Answer:

⚠️ please cannot copy like these!

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let's simplify this step-by-step.

[tex]\[

\lim_{{x \to \infty}} \frac{{3x^3 + 2x^2 + x + 1}}{{x^3 + 2x + 5}}

\][/tex]

### Step 1: Identify the Dominant Terms

[tex]- In \: the \: numerator: {\( 3x^3 \)}[/tex]

[tex]- In \: the \: denominator: {\( x^3 \)}[/tex]

### Step 2: Simplify by Dividing by the Highest Power of \( x \)

Divide each term by \( x^3 \):

[tex]\[

\frac{{3x^3/x^3 + 2x^2/x^3 + x/x^3 + 1/x^3}}{{x^3/x^3 + 2x/x^3 + 5/x^3}}

\][/tex]

Simplify:

[tex]\[

\frac{{3 + \frac{2}{x} + \frac{1}{x^2} + \frac{1}{x^3}}}{{1 + \frac{2}{x^2} + \frac{5}{x^3}}}

\][/tex]

### Step 3: Evaluate the Limit as

[tex]{\( x \to \infty \) As \\( x \) \ approaches \ infinity, \\(\frac{2}{x}\), \(\frac{1}{x^2}\), and \(\frac{1}{x^3}\) \all \approach \0:}[/tex]

[tex]\[

\lim_{{x \to \infty}} \frac{{3 + 0 + 0 + 0}}{{1 + 0 + 0}} = \frac{3}{1} = 3

\][/tex]

So, the limit is:

[tex] \: \[

\boxed{3}

\][/tex]