Mixing
Solving with L'Hôpital's rule. What is wrong?
$begingroup$
L'Hôpital's rule can be used infinitely many times as the limit remains $0/0$. However, this problem does not work with L'Hôpital's rule (I may have counted wrong).
Here it is:
$$lim _{xto 0}left(frac{left(e^x+sin xright)x+e^x-cos x}{x^2}right)$$
We use L'Hôpital's rule once:
$$lim _{xto 0}left(frac{2sin xleft(xright)+xleft(cosleft(xright)+e^xright)+2e^x}{2x}right)$$
One more time:
$$left(frac{−xsinleft(xright)+3cosleft(xright)+left(x+3right)e^x}{2}right)$$
This will equal to $3$. But the real answer is $1$:
$$lim _{xto 0}left(frac{left(e^x+sin xright)x+e^x-cos x}{x^2}right)=1$$
$3$ is not equal to $1$
Thoughts?
calculus limits
edited Jan 13 at 16:26
Robert Z
97.2k1066137
asked Jan 12 at 19:54
Jonathan LarssonJonathan Larsson
83
$endgroup$
|
show 1 more comment
$begingroup$
L'Hôpital's rule can be used infinitely many times as the limit remains $0/0$. However, this problem does not work with L'Hôpital's rule (I may have counted wrong).
Here it is:
$$lim _{xto 0}left(frac{left(e^x+sin xright)x+e^x-cos x}{x^2}right)$$
We use L'Hôpital's rule once:
$$lim _{xto 0}left(frac{2sin xleft(xright)+xleft(cosleft(xright)+e^xright)+2e^x}{2x}right)$$
One more time:
$$left(frac{−xsinleft(xright)+3cosleft(xright)+left(x+3right)e^x}{2}right)$$
This will equal to $3$. But the real answer is $1$:
$$lim _{xto 0}left(frac{left(e^x+sin xright)x+e^x-cos x}{x^2}right)=1$$
$3$ is not equal to $1$
Thoughts?
calculus limits
edited Jan 13 at 16:26
Robert Z
97.2k1066137
asked Jan 12 at 19:54
Jonathan LarssonJonathan Larsson
83
$endgroup$
2
$begingroup$
When doing MathJax, you need to put the maths between dollar signs.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 19:56
$begingroup$
$6$ is not equal to $1$, but $3$ is also not equal to $6$. So is it $6$ or $3$?
$endgroup$
– Dietrich Burde
Jan 12 at 20:00
1
$begingroup$
I don't think you differentiated the first numerator correctly.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 20:01
$begingroup$
Please check the statement of your limit. Is it correct?
$endgroup$
– Robert Z
Jan 12 at 20:07
1
$begingroup$
For me, what you posted tends to $infty$.
$endgroup$
– Bernard
Jan 12 at 20:09
|
show 1 more comment
$begingroup$
L'Hôpital's rule can be used infinitely many times as the limit remains $0/0$. However, this problem does not work with L'Hôpital's rule (I may have counted wrong).
Here it is:
$$lim _{xto 0}left(frac{left(e^x+sin xright)x+e^x-cos x}{x^2}right)$$
We use L'Hôpital's rule once:
$$lim _{xto 0}left(frac{2sin xleft(xright)+xleft(cosleft(xright)+e^xright)+2e^x}{2x}right)$$
One more time:
$$left(frac{−xsinleft(xright)+3cosleft(xright)+left(x+3right)e^x}{2}right)$$
This will equal to $3$. But the real answer is $1$:
$$lim _{xto 0}left(frac{left(e^x+sin xright)x+e^x-cos x}{x^2}right)=1$$
$3$ is not equal to $1$
Thoughts?
calculus limits
edited Jan 13 at 16:26
Robert Z
97.2k1066137
asked Jan 12 at 19:54
Jonathan LarssonJonathan Larsson
83
$endgroup$
L'Hôpital's rule can be used infinitely many times as the limit remains $0/0$. However, this problem does not work with L'Hôpital's rule (I may have counted wrong).
Here it is:
$$lim _{xto 0}left(frac{left(e^x+sin xright)x+e^x-cos x}{x^2}right)$$
We use L'Hôpital's rule once:
$$lim _{xto 0}left(frac{2sin xleft(xright)+xleft(cosleft(xright)+e^xright)+2e^x}{2x}right)$$
One more time:
$$left(frac{−xsinleft(xright)+3cosleft(xright)+left(x+3right)e^x}{2}right)$$
This will equal to $3$. But the real answer is $1$:
$$lim _{xto 0}left(frac{left(e^x+sin xright)x+e^x-cos x}{x^2}right)=1$$
$3$ is not equal to $1$
Thoughts?
calculus limits
calculus limits
edited Jan 13 at 16:26
Robert Z
97.2k1066137
asked Jan 12 at 19:54
Jonathan LarssonJonathan Larsson
83
edited Jan 13 at 16:26
Robert Z
97.2k1066137
asked Jan 12 at 19:54
Jonathan LarssonJonathan Larsson
83
edited Jan 13 at 16:26
Robert Z
97.2k1066137
edited Jan 13 at 16:26
Robert Z
97.2k1066137
edited Jan 13 at 16:26
Robert Z
97.2k1066137
97.2k1066137
asked Jan 12 at 19:54
Jonathan LarssonJonathan Larsson
83
asked Jan 12 at 19:54
Jonathan LarssonJonathan Larsson
83
asked Jan 12 at 19:54
Jonathan LarssonJonathan Larsson
83
83
2
$begingroup$
When doing MathJax, you need to put the maths between dollar signs.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 19:56
$begingroup$
$6$ is not equal to $1$, but $3$ is also not equal to $6$. So is it $6$ or $3$?
$endgroup$
– Dietrich Burde
Jan 12 at 20:00
1
$begingroup$
I don't think you differentiated the first numerator correctly.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 20:01
$begingroup$
Please check the statement of your limit. Is it correct?
$endgroup$
– Robert Z
Jan 12 at 20:07
1
$begingroup$
For me, what you posted tends to $infty$.
$endgroup$
– Bernard
Jan 12 at 20:09
|
show 1 more comment
2
$begingroup$
When doing MathJax, you need to put the maths between dollar signs.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 19:56
$begingroup$
$6$ is not equal to $1$, but $3$ is also not equal to $6$. So is it $6$ or $3$?
$endgroup$
– Dietrich Burde
Jan 12 at 20:00
1
$begingroup$
I don't think you differentiated the first numerator correctly.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 20:01
$begingroup$
Please check the statement of your limit. Is it correct?
$endgroup$
– Robert Z
Jan 12 at 20:07
1
$begingroup$
For me, what you posted tends to $infty$.
$endgroup$
– Bernard
Jan 12 at 20:09
2
2
$begingroup$
When doing MathJax, you need to put the maths between dollar signs.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 19:56
$begingroup$
When doing MathJax, you need to put the maths between dollar signs.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 19:56
$begingroup$
$6$ is not equal to $1$, but $3$ is also not equal to $6$. So is it $6$ or $3$?
$endgroup$
– Dietrich Burde
Jan 12 at 20:00
$begingroup$
$6$ is not equal to $1$, but $3$ is also not equal to $6$. So is it $6$ or $3$?
$endgroup$
– Dietrich Burde
Jan 12 at 20:00
1
1
$begingroup$
I don't think you differentiated the first numerator correctly.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 20:01
$begingroup$
I don't think you differentiated the first numerator correctly.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 20:01
$begingroup$
Please check the statement of your limit. Is it correct?
$endgroup$
– Robert Z
Jan 12 at 20:07
$begingroup$
Please check the statement of your limit. Is it correct?
$endgroup$
– Robert Z
Jan 12 at 20:07
1
1
$begingroup$
For me, what you posted tends to $infty$.
$endgroup$
– Bernard
Jan 12 at 20:09
$begingroup$
For me, what you posted tends to $infty$.
$endgroup$
– Bernard
Jan 12 at 20:09
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
After applying L-Hopital once, the expression that you obtain is not of the indeterminate form ( 0/0 or infinity/infinity) , hence you cannot apply l- hopital again.
Now as after applying l hopital you do not get a finite limit, it does not imply that actual limit does not exist, its just that L-Hopital method fails here.
Important - If you use l'Hopital rule and find that the limit does not exist you cannot conclude that the initial limit does not exist. In that case you must use other methods to analyse the limit.
Note - plotting its graph suggests that its limit does not exist at 0 . It shoots towards infinity.
Also note that by using L'Hopital method you cannot conclude the existence of limit. That's why i used this graphical method of analysis.
answered Jan 12 at 20:05
Swapnil KothriwalSwapnil Kothriwal
513
$endgroup$
add a comment |
$begingroup$
You can't apply l'Hôpital on a non indeterminate form.
Let's see what happens with a Taylor expansion at degree $2$:
begin{align}
(e^x+sin x)x+e^x-cos x
&=(1+x+x+o(x))x+1+x+frac{x^2}{2}-1+frac{x^2}{2}+o(x^2) \
&=x+2x^2+x+x^2+o(x^2)\
&=2x+3x^2+o(x^2)
end{align}
Thus we see that the given limit cannot be finite (and indeed it is $-infty$ from the left and $infty$ from the right).
It's also difficult to understand how the given solution could be $1$, unless we do a sign switch:
begin{align}
(e^x+sin x)x-e^x+cos x
&=(1+x+x+o(x))x-1-x-frac{x^2}{2}+1-frac{x^2}{2}+o(x^2) \
&=x+2x^2-x-x^2+o(x^2)\
&=x^2+o(x^2)
end{align}
This proves that
$$
lim_{xto0}frac{(e^x+sin x)x-e^x+cos x}{x^2}=1
$$
answered Jan 12 at 22:06
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
You can't use L^Hopital's rule since $$lim _{xto 0}left({2sin xleft(xright)+xleft(cosleft(xright)+e^xright)+2e^x}right)=2ne 0$$Also the limit doesn't exist since it is $Large{2over pm 0}$ .
answered Jan 12 at 20:00
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
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votes
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oldest
votes
$begingroup$
After applying L-Hopital once, the expression that you obtain is not of the indeterminate form ( 0/0 or infinity/infinity) , hence you cannot apply l- hopital again.
Now as after applying l hopital you do not get a finite limit, it does not imply that actual limit does not exist, its just that L-Hopital method fails here.
Important - If you use l'Hopital rule and find that the limit does not exist you cannot conclude that the initial limit does not exist. In that case you must use other methods to analyse the limit.
Note - plotting its graph suggests that its limit does not exist at 0 . It shoots towards infinity.
Also note that by using L'Hopital method you cannot conclude the existence of limit. That's why i used this graphical method of analysis.
answered Jan 12 at 20:05
Swapnil KothriwalSwapnil Kothriwal
513
$endgroup$
add a comment |
$begingroup$
After applying L-Hopital once, the expression that you obtain is not of the indeterminate form ( 0/0 or infinity/infinity) , hence you cannot apply l- hopital again.
Now as after applying l hopital you do not get a finite limit, it does not imply that actual limit does not exist, its just that L-Hopital method fails here.
Important - If you use l'Hopital rule and find that the limit does not exist you cannot conclude that the initial limit does not exist. In that case you must use other methods to analyse the limit.
Note - plotting its graph suggests that its limit does not exist at 0 . It shoots towards infinity.
Also note that by using L'Hopital method you cannot conclude the existence of limit. That's why i used this graphical method of analysis.
answered Jan 12 at 20:05
Swapnil KothriwalSwapnil Kothriwal
513
$endgroup$
add a comment |
$begingroup$
After applying L-Hopital once, the expression that you obtain is not of the indeterminate form ( 0/0 or infinity/infinity) , hence you cannot apply l- hopital again.
Now as after applying l hopital you do not get a finite limit, it does not imply that actual limit does not exist, its just that L-Hopital method fails here.
Important - If you use l'Hopital rule and find that the limit does not exist you cannot conclude that the initial limit does not exist. In that case you must use other methods to analyse the limit.
Note - plotting its graph suggests that its limit does not exist at 0 . It shoots towards infinity.
Also note that by using L'Hopital method you cannot conclude the existence of limit. That's why i used this graphical method of analysis.
answered Jan 12 at 20:05
Swapnil KothriwalSwapnil Kothriwal
513
$endgroup$
After applying L-Hopital once, the expression that you obtain is not of the indeterminate form ( 0/0 or infinity/infinity) , hence you cannot apply l- hopital again.
Now as after applying l hopital you do not get a finite limit, it does not imply that actual limit does not exist, its just that L-Hopital method fails here.
Important - If you use l'Hopital rule and find that the limit does not exist you cannot conclude that the initial limit does not exist. In that case you must use other methods to analyse the limit.
Note - plotting its graph suggests that its limit does not exist at 0 . It shoots towards infinity.
Also note that by using L'Hopital method you cannot conclude the existence of limit. That's why i used this graphical method of analysis.
answered Jan 12 at 20:05
Swapnil KothriwalSwapnil Kothriwal
513
edited Jan 12 at 20:31
answered Jan 12 at 20:05
Swapnil KothriwalSwapnil Kothriwal
513
answered Jan 12 at 20:05
Swapnil KothriwalSwapnil Kothriwal
513
answered Jan 12 at 20:05
Swapnil KothriwalSwapnil Kothriwal
513
513
add a comment |
add a comment |
$begingroup$
You can't apply l'Hôpital on a non indeterminate form.
Let's see what happens with a Taylor expansion at degree $2$:
begin{align}
(e^x+sin x)x+e^x-cos x
&=(1+x+x+o(x))x+1+x+frac{x^2}{2}-1+frac{x^2}{2}+o(x^2) \
&=x+2x^2+x+x^2+o(x^2)\
&=2x+3x^2+o(x^2)
end{align}
Thus we see that the given limit cannot be finite (and indeed it is $-infty$ from the left and $infty$ from the right).
It's also difficult to understand how the given solution could be $1$, unless we do a sign switch:
begin{align}
(e^x+sin x)x-e^x+cos x
&=(1+x+x+o(x))x-1-x-frac{x^2}{2}+1-frac{x^2}{2}+o(x^2) \
&=x+2x^2-x-x^2+o(x^2)\
&=x^2+o(x^2)
end{align}
This proves that
$$
lim_{xto0}frac{(e^x+sin x)x-e^x+cos x}{x^2}=1
$$
answered Jan 12 at 22:06
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
You can't apply l'Hôpital on a non indeterminate form.
Let's see what happens with a Taylor expansion at degree $2$:
begin{align}
(e^x+sin x)x+e^x-cos x
&=(1+x+x+o(x))x+1+x+frac{x^2}{2}-1+frac{x^2}{2}+o(x^2) \
&=x+2x^2+x+x^2+o(x^2)\
&=2x+3x^2+o(x^2)
end{align}
Thus we see that the given limit cannot be finite (and indeed it is $-infty$ from the left and $infty$ from the right).
It's also difficult to understand how the given solution could be $1$, unless we do a sign switch:
begin{align}
(e^x+sin x)x-e^x+cos x
&=(1+x+x+o(x))x-1-x-frac{x^2}{2}+1-frac{x^2}{2}+o(x^2) \
&=x+2x^2-x-x^2+o(x^2)\
&=x^2+o(x^2)
end{align}
This proves that
$$
lim_{xto0}frac{(e^x+sin x)x-e^x+cos x}{x^2}=1
$$
answered Jan 12 at 22:06
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
You can't apply l'Hôpital on a non indeterminate form.
Let's see what happens with a Taylor expansion at degree $2$:
begin{align}
(e^x+sin x)x+e^x-cos x
&=(1+x+x+o(x))x+1+x+frac{x^2}{2}-1+frac{x^2}{2}+o(x^2) \
&=x+2x^2+x+x^2+o(x^2)\
&=2x+3x^2+o(x^2)
end{align}
Thus we see that the given limit cannot be finite (and indeed it is $-infty$ from the left and $infty$ from the right).
It's also difficult to understand how the given solution could be $1$, unless we do a sign switch:
begin{align}
(e^x+sin x)x-e^x+cos x
&=(1+x+x+o(x))x-1-x-frac{x^2}{2}+1-frac{x^2}{2}+o(x^2) \
&=x+2x^2-x-x^2+o(x^2)\
&=x^2+o(x^2)
end{align}
This proves that
$$
lim_{xto0}frac{(e^x+sin x)x-e^x+cos x}{x^2}=1
$$
answered Jan 12 at 22:06
egregegreg
181k1485203
$endgroup$
You can't apply l'Hôpital on a non indeterminate form.
Let's see what happens with a Taylor expansion at degree $2$:
begin{align}
(e^x+sin x)x+e^x-cos x
&=(1+x+x+o(x))x+1+x+frac{x^2}{2}-1+frac{x^2}{2}+o(x^2) \
&=x+2x^2+x+x^2+o(x^2)\
&=2x+3x^2+o(x^2)
end{align}
Thus we see that the given limit cannot be finite (and indeed it is $-infty$ from the left and $infty$ from the right).
It's also difficult to understand how the given solution could be $1$, unless we do a sign switch:
begin{align}
(e^x+sin x)x-e^x+cos x
&=(1+x+x+o(x))x-1-x-frac{x^2}{2}+1-frac{x^2}{2}+o(x^2) \
&=x+2x^2-x-x^2+o(x^2)\
&=x^2+o(x^2)
end{align}
This proves that
$$
lim_{xto0}frac{(e^x+sin x)x-e^x+cos x}{x^2}=1
$$
answered Jan 12 at 22:06
egregegreg
181k1485203
answered Jan 12 at 22:06
egregegreg
181k1485203
answered Jan 12 at 22:06
egregegreg
181k1485203
answered Jan 12 at 22:06
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
$begingroup$
You can't use L^Hopital's rule since $$lim _{xto 0}left({2sin xleft(xright)+xleft(cosleft(xright)+e^xright)+2e^x}right)=2ne 0$$Also the limit doesn't exist since it is $Large{2over pm 0}$ .
answered Jan 12 at 20:00
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
You can't use L^Hopital's rule since $$lim _{xto 0}left({2sin xleft(xright)+xleft(cosleft(xright)+e^xright)+2e^x}right)=2ne 0$$Also the limit doesn't exist since it is $Large{2over pm 0}$ .
answered Jan 12 at 20:00
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
You can't use L^Hopital's rule since $$lim _{xto 0}left({2sin xleft(xright)+xleft(cosleft(xright)+e^xright)+2e^x}right)=2ne 0$$Also the limit doesn't exist since it is $Large{2over pm 0}$ .
answered Jan 12 at 20:00
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
You can't use L^Hopital's rule since $$lim _{xto 0}left({2sin xleft(xright)+xleft(cosleft(xright)+e^xright)+2e^x}right)=2ne 0$$Also the limit doesn't exist since it is $Large{2over pm 0}$ .
answered Jan 12 at 20:00
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 20:00
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 20:00
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 20:00
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
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2
$begingroup$
When doing MathJax, you need to put the maths between dollar signs.
$endgroup$
– Lord Shark the Unknown
Jan 12 at 19:56
$begingroup$
$6$ is not equal to $1$, but $3$ is also not equal to $6$. So is it $6$ or $3$?
$endgroup$
– Dietrich Burde
Jan 12 at 20:00
1
$begingroup$
I don't think you differentiated the first numerator correctly.
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– Lord Shark the Unknown
Jan 12 at 20:01
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Please check the statement of your limit. Is it correct?
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– Robert Z
Jan 12 at 20:07
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For me, what you posted tends to $infty$.
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– Bernard
Jan 12 at 20:09