Proving Limit Rigorously
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1
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Find the limit
$$large lim_{xto infty}(ln x)^{frac{20}x}$$
I understood that as x approached infinity, $20/x$ approached 0. This would mean that the limit would tend toward $1$. However, $ln x$ also approaches infinity as $x$ approaches infinity. Thus, I suspected the answer to be $1$ (and it indeed is the answer), however I feel like this answer is not sufficiently rigorous. How could I rigorously prove $1$ as the answer? Any ideas/hints would be appreciated.
Note: I am a highschooler and my teachers often tell me to take these answers in faith. Thus, I may not understand any fancy notations that may usually be used in solving limit.
Thanks!
calculus limits
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up vote
1
down vote
favorite
Find the limit
$$large lim_{xto infty}(ln x)^{frac{20}x}$$
I understood that as x approached infinity, $20/x$ approached 0. This would mean that the limit would tend toward $1$. However, $ln x$ also approaches infinity as $x$ approaches infinity. Thus, I suspected the answer to be $1$ (and it indeed is the answer), however I feel like this answer is not sufficiently rigorous. How could I rigorously prove $1$ as the answer? Any ideas/hints would be appreciated.
Note: I am a highschooler and my teachers often tell me to take these answers in faith. Thus, I may not understand any fancy notations that may usually be used in solving limit.
Thanks!
calculus limits
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find the limit
$$large lim_{xto infty}(ln x)^{frac{20}x}$$
I understood that as x approached infinity, $20/x$ approached 0. This would mean that the limit would tend toward $1$. However, $ln x$ also approaches infinity as $x$ approaches infinity. Thus, I suspected the answer to be $1$ (and it indeed is the answer), however I feel like this answer is not sufficiently rigorous. How could I rigorously prove $1$ as the answer? Any ideas/hints would be appreciated.
Note: I am a highschooler and my teachers often tell me to take these answers in faith. Thus, I may not understand any fancy notations that may usually be used in solving limit.
Thanks!
calculus limits
Find the limit
$$large lim_{xto infty}(ln x)^{frac{20}x}$$
I understood that as x approached infinity, $20/x$ approached 0. This would mean that the limit would tend toward $1$. However, $ln x$ also approaches infinity as $x$ approaches infinity. Thus, I suspected the answer to be $1$ (and it indeed is the answer), however I feel like this answer is not sufficiently rigorous. How could I rigorously prove $1$ as the answer? Any ideas/hints would be appreciated.
Note: I am a highschooler and my teachers often tell me to take these answers in faith. Thus, I may not understand any fancy notations that may usually be used in solving limit.
Thanks!
calculus limits
calculus limits
edited 58 mins ago
gimusi
87.1k74393
87.1k74393
asked 1 hour ago
Dude156
493214
493214
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add a comment |
3 Answers
3
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oldest
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up vote
4
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The argument you use lacks the necessary rigour. With the same argument, you would conclude that $bigl(mathrm e^xbigr)^tfrac1xto 1$, yet $bigl(mathrm e^xbigr)^tfrac1x$ is Euler's number $mathrm e$!
The below solutions are fine. However, as a highschooler, you might want a rigourous proof that $frac{ln(ln x)}{x}$ tends to $0$ as $x$ tends to $infty$. Here's a sketch of a simple proof:
$$frac{ln(ln x)}{x}=underbrace{frac{ln(ln x)}{ln x}}_{begin{matrix}downarrow\0\text{(setting }u=ln x)end{matrix}}!!underbrace{frac{ln x}{x}}_{begin{matrix}downarrow\0end{matrix}} $$
Thanks for the reply Bernard! Can you please follow through with that proof? I just used L Hospitals, but I'd be interested to see an alternate method.
– Dude156
12 mins ago
add a comment |
up vote
2
down vote
First compute the limit of the logarithm of the expression. Indeed,
$$
frac{20}{x}times ln(ln x)to 0
$$
as $xto infty$. You can see this intuitively since $x$ grows much faster than $ln(ln x)$ or you can use L'Hospital's rule. In any case
$$
expleft(frac{20}{x}times ln(ln x)right)to 1
$$
as $xto infty$
add a comment |
up vote
2
down vote
HINT
Use that
$$large (ln x)^{frac{20}x}= e^{left[20frac{ln(ln x)}{x}right]}$$
or as an alternative $$large (ln x)^{frac{20}x}=left[(ln x)^{1/ln x}right]^{frac{20ln x}x}$$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
The argument you use lacks the necessary rigour. With the same argument, you would conclude that $bigl(mathrm e^xbigr)^tfrac1xto 1$, yet $bigl(mathrm e^xbigr)^tfrac1x$ is Euler's number $mathrm e$!
The below solutions are fine. However, as a highschooler, you might want a rigourous proof that $frac{ln(ln x)}{x}$ tends to $0$ as $x$ tends to $infty$. Here's a sketch of a simple proof:
$$frac{ln(ln x)}{x}=underbrace{frac{ln(ln x)}{ln x}}_{begin{matrix}downarrow\0\text{(setting }u=ln x)end{matrix}}!!underbrace{frac{ln x}{x}}_{begin{matrix}downarrow\0end{matrix}} $$
Thanks for the reply Bernard! Can you please follow through with that proof? I just used L Hospitals, but I'd be interested to see an alternate method.
– Dude156
12 mins ago
add a comment |
up vote
4
down vote
The argument you use lacks the necessary rigour. With the same argument, you would conclude that $bigl(mathrm e^xbigr)^tfrac1xto 1$, yet $bigl(mathrm e^xbigr)^tfrac1x$ is Euler's number $mathrm e$!
The below solutions are fine. However, as a highschooler, you might want a rigourous proof that $frac{ln(ln x)}{x}$ tends to $0$ as $x$ tends to $infty$. Here's a sketch of a simple proof:
$$frac{ln(ln x)}{x}=underbrace{frac{ln(ln x)}{ln x}}_{begin{matrix}downarrow\0\text{(setting }u=ln x)end{matrix}}!!underbrace{frac{ln x}{x}}_{begin{matrix}downarrow\0end{matrix}} $$
Thanks for the reply Bernard! Can you please follow through with that proof? I just used L Hospitals, but I'd be interested to see an alternate method.
– Dude156
12 mins ago
add a comment |
up vote
4
down vote
up vote
4
down vote
The argument you use lacks the necessary rigour. With the same argument, you would conclude that $bigl(mathrm e^xbigr)^tfrac1xto 1$, yet $bigl(mathrm e^xbigr)^tfrac1x$ is Euler's number $mathrm e$!
The below solutions are fine. However, as a highschooler, you might want a rigourous proof that $frac{ln(ln x)}{x}$ tends to $0$ as $x$ tends to $infty$. Here's a sketch of a simple proof:
$$frac{ln(ln x)}{x}=underbrace{frac{ln(ln x)}{ln x}}_{begin{matrix}downarrow\0\text{(setting }u=ln x)end{matrix}}!!underbrace{frac{ln x}{x}}_{begin{matrix}downarrow\0end{matrix}} $$
The argument you use lacks the necessary rigour. With the same argument, you would conclude that $bigl(mathrm e^xbigr)^tfrac1xto 1$, yet $bigl(mathrm e^xbigr)^tfrac1x$ is Euler's number $mathrm e$!
The below solutions are fine. However, as a highschooler, you might want a rigourous proof that $frac{ln(ln x)}{x}$ tends to $0$ as $x$ tends to $infty$. Here's a sketch of a simple proof:
$$frac{ln(ln x)}{x}=underbrace{frac{ln(ln x)}{ln x}}_{begin{matrix}downarrow\0\text{(setting }u=ln x)end{matrix}}!!underbrace{frac{ln x}{x}}_{begin{matrix}downarrow\0end{matrix}} $$
edited 13 mins ago
Dude156
493214
493214
answered 43 mins ago
Bernard
116k637107
116k637107
Thanks for the reply Bernard! Can you please follow through with that proof? I just used L Hospitals, but I'd be interested to see an alternate method.
– Dude156
12 mins ago
add a comment |
Thanks for the reply Bernard! Can you please follow through with that proof? I just used L Hospitals, but I'd be interested to see an alternate method.
– Dude156
12 mins ago
Thanks for the reply Bernard! Can you please follow through with that proof? I just used L Hospitals, but I'd be interested to see an alternate method.
– Dude156
12 mins ago
Thanks for the reply Bernard! Can you please follow through with that proof? I just used L Hospitals, but I'd be interested to see an alternate method.
– Dude156
12 mins ago
add a comment |
up vote
2
down vote
First compute the limit of the logarithm of the expression. Indeed,
$$
frac{20}{x}times ln(ln x)to 0
$$
as $xto infty$. You can see this intuitively since $x$ grows much faster than $ln(ln x)$ or you can use L'Hospital's rule. In any case
$$
expleft(frac{20}{x}times ln(ln x)right)to 1
$$
as $xto infty$
add a comment |
up vote
2
down vote
First compute the limit of the logarithm of the expression. Indeed,
$$
frac{20}{x}times ln(ln x)to 0
$$
as $xto infty$. You can see this intuitively since $x$ grows much faster than $ln(ln x)$ or you can use L'Hospital's rule. In any case
$$
expleft(frac{20}{x}times ln(ln x)right)to 1
$$
as $xto infty$
add a comment |
up vote
2
down vote
up vote
2
down vote
First compute the limit of the logarithm of the expression. Indeed,
$$
frac{20}{x}times ln(ln x)to 0
$$
as $xto infty$. You can see this intuitively since $x$ grows much faster than $ln(ln x)$ or you can use L'Hospital's rule. In any case
$$
expleft(frac{20}{x}times ln(ln x)right)to 1
$$
as $xto infty$
First compute the limit of the logarithm of the expression. Indeed,
$$
frac{20}{x}times ln(ln x)to 0
$$
as $xto infty$. You can see this intuitively since $x$ grows much faster than $ln(ln x)$ or you can use L'Hospital's rule. In any case
$$
expleft(frac{20}{x}times ln(ln x)right)to 1
$$
as $xto infty$
edited 1 hour ago
Bernard
116k637107
116k637107
answered 1 hour ago
Foobaz John
19.7k41250
19.7k41250
add a comment |
add a comment |
up vote
2
down vote
HINT
Use that
$$large (ln x)^{frac{20}x}= e^{left[20frac{ln(ln x)}{x}right]}$$
or as an alternative $$large (ln x)^{frac{20}x}=left[(ln x)^{1/ln x}right]^{frac{20ln x}x}$$
add a comment |
up vote
2
down vote
HINT
Use that
$$large (ln x)^{frac{20}x}= e^{left[20frac{ln(ln x)}{x}right]}$$
or as an alternative $$large (ln x)^{frac{20}x}=left[(ln x)^{1/ln x}right]^{frac{20ln x}x}$$
add a comment |
up vote
2
down vote
up vote
2
down vote
HINT
Use that
$$large (ln x)^{frac{20}x}= e^{left[20frac{ln(ln x)}{x}right]}$$
or as an alternative $$large (ln x)^{frac{20}x}=left[(ln x)^{1/ln x}right]^{frac{20ln x}x}$$
HINT
Use that
$$large (ln x)^{frac{20}x}= e^{left[20frac{ln(ln x)}{x}right]}$$
or as an alternative $$large (ln x)^{frac{20}x}=left[(ln x)^{1/ln x}right]^{frac{20ln x}x}$$
edited 59 mins ago
answered 1 hour ago
gimusi
87.1k74393
87.1k74393
add a comment |
add a comment |
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