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How to transform this limit expression as a limit of $e$
up vote
1
down vote
favorite
I have the following expression:
$displaystylelim _{xto infty }left(dfrac{x+2}{:x-6}right)^left(dfrac{x}{:4}right)$
I'm studying Calculus I and our lector has shown us ways of transforming such limits to:
$displaystylelim_{xto infty }left(1+frac{1}{:x}right)^x = e$
The way this calculator solves it is not immediately clear to me, is there any other way to find the above limit?
calculus
edited yesterday
Yadati Kiran
364111
asked yesterday
PowerUser
12518
add a comment |
up vote
1
down vote
favorite
I have the following expression:
$displaystylelim _{xto infty }left(dfrac{x+2}{:x-6}right)^left(dfrac{x}{:4}right)$
I'm studying Calculus I and our lector has shown us ways of transforming such limits to:
$displaystylelim_{xto infty }left(1+frac{1}{:x}right)^x = e$
The way this calculator solves it is not immediately clear to me, is there any other way to find the above limit?
calculus
edited yesterday
Yadati Kiran
364111
asked yesterday
PowerUser
12518
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have the following expression:
$displaystylelim _{xto infty }left(dfrac{x+2}{:x-6}right)^left(dfrac{x}{:4}right)$
I'm studying Calculus I and our lector has shown us ways of transforming such limits to:
$displaystylelim_{xto infty }left(1+frac{1}{:x}right)^x = e$
The way this calculator solves it is not immediately clear to me, is there any other way to find the above limit?
calculus
edited yesterday
Yadati Kiran
364111
asked yesterday
PowerUser
12518
I have the following expression:
$displaystylelim _{xto infty }left(dfrac{x+2}{:x-6}right)^left(dfrac{x}{:4}right)$
I'm studying Calculus I and our lector has shown us ways of transforming such limits to:
$displaystylelim_{xto infty }left(1+frac{1}{:x}right)^x = e$
The way this calculator solves it is not immediately clear to me, is there any other way to find the above limit?
calculus
calculus
edited yesterday
Yadati Kiran
364111
asked yesterday
PowerUser
12518
edited yesterday
Yadati Kiran
364111
asked yesterday
PowerUser
12518
edited yesterday
Yadati Kiran
364111
edited yesterday
Yadati Kiran
364111
edited yesterday
Yadati Kiran
364111
364111
asked yesterday
PowerUser
12518
asked yesterday
PowerUser
12518
asked yesterday
PowerUser
12518
12518
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
HINT
We have that
$$left(frac{x+2}{x-6}right)^{frac x 4}=left(frac{x-6+8}{x-6}right)^{frac x 4}=left(1+frac{8}{x-6}right)^{frac x 4}$$
then we can manipulate further in order to use the standard limit.
Refer to the related
- Calculating a limit with trignonometeric and quadratic function
answered yesterday
gimusi
85.6k74294
I've already reached that, the problem is I don't know what to do next. I tried adding and subtracting 6/4 from the power, but it didn't quite work out...
– PowerUser
yesterday
@PowerUser Use that $A^n=(A^m)^{frac n m}$.
– gimusi
yesterday
I solved it. What I used was A^(n/m) = (A^n)^(1/m) and got e^(8/4) = e^2. Sorry for the bad formatting, I'm not used to MathJax… Seeing as this is the oldest answer I will accept it. Thanks!
– PowerUser
yesterday
@PowerUser You are welcome! Bye
– gimusi
yesterday
@PowerUser Recall that as ana alternative we can also use that $$f(x)^{g(x)}=e^{g(x) log (f(x))}$$ to obtain teh same result by standard limit $$t to 0 quad frac{log (1+t)}t to 1$$
– gimusi
yesterday
add a comment |
up vote
1
down vote
You may proceed as follows:
- Set $y = x-6$
$$left(frac{x+2}{x-6} right)^{frac{x}{4}} = left(1 +frac{8}{y} right)^{frac{y+6}{4}} = left(1 +frac{2}{frac{y}{4}} right)^{frac{y}{4}}cdot left(1 +frac{8}{y} right)^{frac{3}{2}} stackrel{y to infty}{longrightarrow}e^2$$
answered yesterday
trancelocation
8,0561519
add a comment |
up vote
0
down vote
If you divide $x+2$ by $x-6$ the quotient is $1$ and the remainder is $8$, so the limit becomes
$$lim_{xtoinfty}left(1+frac{8}{x-6}right)^{x/4}$$
Now, substitute $x/4$ by
$$frac{x-6}8cdotfrac{8}{x-6}cdotfrac x4$$
to get
$$lim_{xtoinfty}left[left(1+frac{1}{dfrac{x-6}8}right)^{dfrac{x-6}8}right]^{dfrac{8x}{4(x-6)}}$$
answered yesterday
ajotatxe
52.1k23688
add a comment |
up vote
0
down vote
$displaystylelim _{xto infty }left(dfrac{x+2}{:x-6}right)^left(dfrac{x}{:4}right)=lim _{xto infty }left(dfrac{1+dfrac2x}{1-dfrac6x}right)^left(dfrac{x}{:4}right)=dfrac{left(lim_{xtoinfty}left(1+dfrac2xright)^{x/2}right)^{1/2}}{left(lim_{xtoinfty}left(1-dfrac6xright)^{-x/6}right)^{-3/2}}=dfrac{e^{1/2}}{e^{-3/2}}=?$
answered yesterday
lab bhattacharjee
219k15154270
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
HINT
We have that
$$left(frac{x+2}{x-6}right)^{frac x 4}=left(frac{x-6+8}{x-6}right)^{frac x 4}=left(1+frac{8}{x-6}right)^{frac x 4}$$
then we can manipulate further in order to use the standard limit.
Refer to the related
- Calculating a limit with trignonometeric and quadratic function
answered yesterday
gimusi
85.6k74294
I've already reached that, the problem is I don't know what to do next. I tried adding and subtracting 6/4 from the power, but it didn't quite work out...
– PowerUser
yesterday
@PowerUser Use that $A^n=(A^m)^{frac n m}$.
– gimusi
yesterday
I solved it. What I used was A^(n/m) = (A^n)^(1/m) and got e^(8/4) = e^2. Sorry for the bad formatting, I'm not used to MathJax… Seeing as this is the oldest answer I will accept it. Thanks!
– PowerUser
yesterday
@PowerUser You are welcome! Bye
– gimusi
yesterday
@PowerUser Recall that as ana alternative we can also use that $$f(x)^{g(x)}=e^{g(x) log (f(x))}$$ to obtain teh same result by standard limit $$t to 0 quad frac{log (1+t)}t to 1$$
– gimusi
yesterday
add a comment |
up vote
3
down vote
accepted
HINT
We have that
$$left(frac{x+2}{x-6}right)^{frac x 4}=left(frac{x-6+8}{x-6}right)^{frac x 4}=left(1+frac{8}{x-6}right)^{frac x 4}$$
then we can manipulate further in order to use the standard limit.
Refer to the related
- Calculating a limit with trignonometeric and quadratic function
answered yesterday
gimusi
85.6k74294
I've already reached that, the problem is I don't know what to do next. I tried adding and subtracting 6/4 from the power, but it didn't quite work out...
– PowerUser
yesterday
@PowerUser Use that $A^n=(A^m)^{frac n m}$.
– gimusi
yesterday
I solved it. What I used was A^(n/m) = (A^n)^(1/m) and got e^(8/4) = e^2. Sorry for the bad formatting, I'm not used to MathJax… Seeing as this is the oldest answer I will accept it. Thanks!
– PowerUser
yesterday
@PowerUser You are welcome! Bye
– gimusi
yesterday
@PowerUser Recall that as ana alternative we can also use that $$f(x)^{g(x)}=e^{g(x) log (f(x))}$$ to obtain teh same result by standard limit $$t to 0 quad frac{log (1+t)}t to 1$$
– gimusi
yesterday
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
HINT
We have that
$$left(frac{x+2}{x-6}right)^{frac x 4}=left(frac{x-6+8}{x-6}right)^{frac x 4}=left(1+frac{8}{x-6}right)^{frac x 4}$$
then we can manipulate further in order to use the standard limit.
Refer to the related
- Calculating a limit with trignonometeric and quadratic function
answered yesterday
gimusi
85.6k74294
HINT
We have that
$$left(frac{x+2}{x-6}right)^{frac x 4}=left(frac{x-6+8}{x-6}right)^{frac x 4}=left(1+frac{8}{x-6}right)^{frac x 4}$$
then we can manipulate further in order to use the standard limit.
Refer to the related
- Calculating a limit with trignonometeric and quadratic function
answered yesterday
gimusi
85.6k74294
edited yesterday
answered yesterday
gimusi
85.6k74294
answered yesterday
gimusi
85.6k74294
answered yesterday
gimusi
85.6k74294
85.6k74294
I've already reached that, the problem is I don't know what to do next. I tried adding and subtracting 6/4 from the power, but it didn't quite work out...
– PowerUser
yesterday
@PowerUser Use that $A^n=(A^m)^{frac n m}$.
– gimusi
yesterday
I solved it. What I used was A^(n/m) = (A^n)^(1/m) and got e^(8/4) = e^2. Sorry for the bad formatting, I'm not used to MathJax… Seeing as this is the oldest answer I will accept it. Thanks!
– PowerUser
yesterday
@PowerUser You are welcome! Bye
– gimusi
yesterday
@PowerUser Recall that as ana alternative we can also use that $$f(x)^{g(x)}=e^{g(x) log (f(x))}$$ to obtain teh same result by standard limit $$t to 0 quad frac{log (1+t)}t to 1$$
– gimusi
yesterday
add a comment |
I've already reached that, the problem is I don't know what to do next. I tried adding and subtracting 6/4 from the power, but it didn't quite work out...
– PowerUser
yesterday
@PowerUser Use that $A^n=(A^m)^{frac n m}$.
– gimusi
yesterday
I solved it. What I used was A^(n/m) = (A^n)^(1/m) and got e^(8/4) = e^2. Sorry for the bad formatting, I'm not used to MathJax… Seeing as this is the oldest answer I will accept it. Thanks!
– PowerUser
yesterday
@PowerUser You are welcome! Bye
– gimusi
yesterday
@PowerUser Recall that as ana alternative we can also use that $$f(x)^{g(x)}=e^{g(x) log (f(x))}$$ to obtain teh same result by standard limit $$t to 0 quad frac{log (1+t)}t to 1$$
– gimusi
yesterday
I've already reached that, the problem is I don't know what to do next. I tried adding and subtracting 6/4 from the power, but it didn't quite work out...
– PowerUser
yesterday
I've already reached that, the problem is I don't know what to do next. I tried adding and subtracting 6/4 from the power, but it didn't quite work out...
– PowerUser
yesterday
@PowerUser Use that $A^n=(A^m)^{frac n m}$.
– gimusi
yesterday
@PowerUser Use that $A^n=(A^m)^{frac n m}$.
– gimusi
yesterday
I solved it. What I used was A^(n/m) = (A^n)^(1/m) and got e^(8/4) = e^2. Sorry for the bad formatting, I'm not used to MathJax… Seeing as this is the oldest answer I will accept it. Thanks!
– PowerUser
yesterday
I solved it. What I used was A^(n/m) = (A^n)^(1/m) and got e^(8/4) = e^2. Sorry for the bad formatting, I'm not used to MathJax… Seeing as this is the oldest answer I will accept it. Thanks!
– PowerUser
yesterday
@PowerUser You are welcome! Bye
– gimusi
yesterday
@PowerUser You are welcome! Bye
– gimusi
yesterday
@PowerUser Recall that as ana alternative we can also use that $$f(x)^{g(x)}=e^{g(x) log (f(x))}$$ to obtain teh same result by standard limit $$t to 0 quad frac{log (1+t)}t to 1$$
– gimusi
yesterday
@PowerUser Recall that as ana alternative we can also use that $$f(x)^{g(x)}=e^{g(x) log (f(x))}$$ to obtain teh same result by standard limit $$t to 0 quad frac{log (1+t)}t to 1$$
– gimusi
yesterday
add a comment |
up vote
1
down vote
You may proceed as follows:
- Set $y = x-6$
$$left(frac{x+2}{x-6} right)^{frac{x}{4}} = left(1 +frac{8}{y} right)^{frac{y+6}{4}} = left(1 +frac{2}{frac{y}{4}} right)^{frac{y}{4}}cdot left(1 +frac{8}{y} right)^{frac{3}{2}} stackrel{y to infty}{longrightarrow}e^2$$
answered yesterday
trancelocation
8,0561519
add a comment |
up vote
1
down vote
You may proceed as follows:
- Set $y = x-6$
$$left(frac{x+2}{x-6} right)^{frac{x}{4}} = left(1 +frac{8}{y} right)^{frac{y+6}{4}} = left(1 +frac{2}{frac{y}{4}} right)^{frac{y}{4}}cdot left(1 +frac{8}{y} right)^{frac{3}{2}} stackrel{y to infty}{longrightarrow}e^2$$
answered yesterday
trancelocation
8,0561519
add a comment |
up vote
1
down vote
up vote
1
down vote
You may proceed as follows:
- Set $y = x-6$
$$left(frac{x+2}{x-6} right)^{frac{x}{4}} = left(1 +frac{8}{y} right)^{frac{y+6}{4}} = left(1 +frac{2}{frac{y}{4}} right)^{frac{y}{4}}cdot left(1 +frac{8}{y} right)^{frac{3}{2}} stackrel{y to infty}{longrightarrow}e^2$$
answered yesterday
trancelocation
8,0561519
You may proceed as follows:
- Set $y = x-6$
$$left(frac{x+2}{x-6} right)^{frac{x}{4}} = left(1 +frac{8}{y} right)^{frac{y+6}{4}} = left(1 +frac{2}{frac{y}{4}} right)^{frac{y}{4}}cdot left(1 +frac{8}{y} right)^{frac{3}{2}} stackrel{y to infty}{longrightarrow}e^2$$
answered yesterday
trancelocation
8,0561519
answered yesterday
trancelocation
8,0561519
answered yesterday
trancelocation
8,0561519
answered yesterday
trancelocation
8,0561519
8,0561519
add a comment |
add a comment |
up vote
0
down vote
If you divide $x+2$ by $x-6$ the quotient is $1$ and the remainder is $8$, so the limit becomes
$$lim_{xtoinfty}left(1+frac{8}{x-6}right)^{x/4}$$
Now, substitute $x/4$ by
$$frac{x-6}8cdotfrac{8}{x-6}cdotfrac x4$$
to get
$$lim_{xtoinfty}left[left(1+frac{1}{dfrac{x-6}8}right)^{dfrac{x-6}8}right]^{dfrac{8x}{4(x-6)}}$$
answered yesterday
ajotatxe
52.1k23688
add a comment |
up vote
0
down vote
If you divide $x+2$ by $x-6$ the quotient is $1$ and the remainder is $8$, so the limit becomes
$$lim_{xtoinfty}left(1+frac{8}{x-6}right)^{x/4}$$
Now, substitute $x/4$ by
$$frac{x-6}8cdotfrac{8}{x-6}cdotfrac x4$$
to get
$$lim_{xtoinfty}left[left(1+frac{1}{dfrac{x-6}8}right)^{dfrac{x-6}8}right]^{dfrac{8x}{4(x-6)}}$$
answered yesterday
ajotatxe
52.1k23688
add a comment |
up vote
0
down vote
up vote
0
down vote
If you divide $x+2$ by $x-6$ the quotient is $1$ and the remainder is $8$, so the limit becomes
$$lim_{xtoinfty}left(1+frac{8}{x-6}right)^{x/4}$$
Now, substitute $x/4$ by
$$frac{x-6}8cdotfrac{8}{x-6}cdotfrac x4$$
to get
$$lim_{xtoinfty}left[left(1+frac{1}{dfrac{x-6}8}right)^{dfrac{x-6}8}right]^{dfrac{8x}{4(x-6)}}$$
answered yesterday
ajotatxe
52.1k23688
If you divide $x+2$ by $x-6$ the quotient is $1$ and the remainder is $8$, so the limit becomes
$$lim_{xtoinfty}left(1+frac{8}{x-6}right)^{x/4}$$
Now, substitute $x/4$ by
$$frac{x-6}8cdotfrac{8}{x-6}cdotfrac x4$$
to get
$$lim_{xtoinfty}left[left(1+frac{1}{dfrac{x-6}8}right)^{dfrac{x-6}8}right]^{dfrac{8x}{4(x-6)}}$$
answered yesterday
ajotatxe
52.1k23688
answered yesterday
ajotatxe
52.1k23688
answered yesterday
ajotatxe
52.1k23688
answered yesterday
ajotatxe
52.1k23688
52.1k23688
add a comment |
add a comment |
up vote
0
down vote
$displaystylelim _{xto infty }left(dfrac{x+2}{:x-6}right)^left(dfrac{x}{:4}right)=lim _{xto infty }left(dfrac{1+dfrac2x}{1-dfrac6x}right)^left(dfrac{x}{:4}right)=dfrac{left(lim_{xtoinfty}left(1+dfrac2xright)^{x/2}right)^{1/2}}{left(lim_{xtoinfty}left(1-dfrac6xright)^{-x/6}right)^{-3/2}}=dfrac{e^{1/2}}{e^{-3/2}}=?$
answered yesterday
lab bhattacharjee
219k15154270
add a comment |
up vote
0
down vote
$displaystylelim _{xto infty }left(dfrac{x+2}{:x-6}right)^left(dfrac{x}{:4}right)=lim _{xto infty }left(dfrac{1+dfrac2x}{1-dfrac6x}right)^left(dfrac{x}{:4}right)=dfrac{left(lim_{xtoinfty}left(1+dfrac2xright)^{x/2}right)^{1/2}}{left(lim_{xtoinfty}left(1-dfrac6xright)^{-x/6}right)^{-3/2}}=dfrac{e^{1/2}}{e^{-3/2}}=?$
answered yesterday
lab bhattacharjee
219k15154270
add a comment |
up vote
0
down vote
up vote
0
down vote
$displaystylelim _{xto infty }left(dfrac{x+2}{:x-6}right)^left(dfrac{x}{:4}right)=lim _{xto infty }left(dfrac{1+dfrac2x}{1-dfrac6x}right)^left(dfrac{x}{:4}right)=dfrac{left(lim_{xtoinfty}left(1+dfrac2xright)^{x/2}right)^{1/2}}{left(lim_{xtoinfty}left(1-dfrac6xright)^{-x/6}right)^{-3/2}}=dfrac{e^{1/2}}{e^{-3/2}}=?$
answered yesterday
lab bhattacharjee
219k15154270
$displaystylelim _{xto infty }left(dfrac{x+2}{:x-6}right)^left(dfrac{x}{:4}right)=lim _{xto infty }left(dfrac{1+dfrac2x}{1-dfrac6x}right)^left(dfrac{x}{:4}right)=dfrac{left(lim_{xtoinfty}left(1+dfrac2xright)^{x/2}right)^{1/2}}{left(lim_{xtoinfty}left(1-dfrac6xright)^{-x/6}right)^{-3/2}}=dfrac{e^{1/2}}{e^{-3/2}}=?$
answered yesterday
lab bhattacharjee
219k15154270
answered yesterday
lab bhattacharjee
219k15154270
answered yesterday
lab bhattacharjee
219k15154270
answered yesterday
lab bhattacharjee
219k15154270
219k15154270
add a comment |
add a comment |
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