009C Sample Final 1, Problem 1

Compute

a) $\lim _{n\rightarrow \infty }{\frac {3-2n^{2}}{5n^{2}+n+1}}$

b) $\lim _{n\rightarrow \infty }{\frac {\ln n}{\ln 3n}}$

Foundations:
Review L'Hopital's Rule

Solution:

(a)

Step 1:
First, we switch to the limit to $x$ so that we can use L'Hopital's rule.
So, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\ &&\\ & \overset{l'H}{=} & \displaystyle{\frac{-4}{10}}\\ &&\\ & = & \displaystyle{\frac{-2}{5}} \end{array}}

Step 2:
Hence, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}}

(b)

Step 1:

Step 2:

Step 3:

Final Answer:
(a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-2}{5}}
(b)

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