[tex]\underline{\underline{\large{\red{\mathcal{✒GIVEN:}}}}}[/tex]
[tex]\bullet \: \: \rm{\lim_{{x \to \infty}} \dfrac{{x^2 + 3x + 2}}{{2x^2 + 5x + 3}}}[/tex]
[tex]\underline{\underline{\large{\red{\mathcal{REQUIRED:}}}}}[/tex]
Solve the equation the limit using L'Hospital's rule.
[tex]\underline{\underline{\large{\red{\mathcal{SOLUTION:}}}}}[/tex]
Hi, Brainly User! Let me help you finding the limit ^^
1) Solve the limit below using [tex]\tt{ \purple{L'Hopital's \: rule}}[/tex]:
[tex]\tt{\lim_{{x \to \infty}} \dfrac{{x^2 + 3x + 2}}{{2x^2 + 5x + 3}}}[/tex]
2) We can use L'Hopital's rule if, when evaluated, the limit is of the form 0/0, of +/- inf/inf:
[tex]\tt{Which \: states \: that:}[/tex]
[tex]\small{\boxed{ \bm{{ \red{\lim_{{x \to \infty}} \dfrac{{f(x)}}{{g(x)}} = \lim_{{x \to \infty}} \dfrac{{f'(x)}}{{g'(x)}}}}}}}[/tex]
Therefore, we have:
[tex]\tt{\implies \lim_{{x \to \infty}} \frac{{ {x}^{2} + 3x + 2}}{{ {2x}^{2} + 5x + 3}} = \lim_{{x \to \infty}} \frac{{2x + 3}}{{4x + 5}} }[/tex]
3) Once again, when evaluated, the limit has the form inf/inf; thus, we can use L'Hopital's rule a second time:
[tex]\tt{\implies \lim_{{x \to \infty}} \frac{{ {x}^{2} + 3x + 2}}{{ {2x}^{2} + 5x + 3}} = \lim_{{x \to \infty}} \frac{2}{4} =\large {\purple{ \frac{1}{2}}} }[/tex]
Final Answer:
Hence, the limit is equal to [tex]\rm{\purple{\dfrac{1}{2}}}[/tex].
