009C Sample Final 1, Problem 1
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Compute
a)
lim
n
→
∞
3
−
2
n
2
5
n
2
+
n
+
1
{\displaystyle \lim _{n\rightarrow \infty }{\frac {3-2n^{2}}{5n^{2}+n+1}}}
b)
lim
n
→
∞
ln
n
ln
3
n
{\displaystyle \lim _{n\rightarrow \infty }{\frac {\ln n}{\ln 3n}}}
Foundations:
Recall:
L'Hopital's Rule
Suppose that
lim
x
→
∞
f
(
x
)
{\displaystyle \lim _{x\rightarrow \infty }f(x)}
and
lim
x
→
∞
g
(
x
)
{\displaystyle \lim _{x\rightarrow \infty }g(x)}
are both zero or both
±
∞
.
{\displaystyle \pm \infty .}
If
lim
x
→
∞
f
′
(
x
)
g
′
(
x
)
{\displaystyle \lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}}
is finite or
±
∞
,
{\displaystyle \pm \infty ,}
then
lim
x
→
∞
f
(
x
)
g
(
x
)
=
lim
x
→
∞
f
′
(
x
)
g
′
(
x
)
.
{\displaystyle \lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.}
Solution:
(a)
Step 1:
First, we switch to the limit to
x
{\displaystyle x}
so that we can use L'Hopital's rule.
So, we have
lim
x
→
∞
3
−
2
x
2
5
x
2
+
x
+
1
=
l
′
H
lim
x
→
∞
−
4
x
10
x
+
1
=
l
′
H
−
4
10
=
−
2
5
.
{\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
lim
n
→
∞
3
−
2
n
2
5
n
2
+
n
+
1
=
−
2
5
.
{\displaystyle \lim _{n\rightarrow \infty }{\frac {3-2n^{2}}{5n^{2}+n+1}}={\frac {-2}{5}}.}
(b)
Step 1:
Again, we switch to the limit to
x
{\displaystyle x}
so that we can use L'Hopital's rule.
So, we have
lim
x
→
∞
ln
x
ln
(
3
x
)
=
l
′
H
lim
x
→
∞
1
x
3
3
x
=
lim
x
→
∞
1
=
1.
{\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln x}{\ln(3x)}}}&{\overset {l'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\frac {1}{x}}{\frac {3}{3x}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }1}\\&&\\&=&1.\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{\ln n}{\ln 3n}=1.}
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)
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 1}
Return to Sample Exam
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