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What to do when direct substitution tells us that limit does not exist but in actual it does exist?
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I am learning limits these days and I have reached at finding limits by direct substitution but there's one that doesn't making sense to me at all. Let me illustrate it with an example $f(x) = x/x$ .The limit ( as $x$ approaches $0$) in this case does exist but by direct substitution, it says limit does not exist. What to do when you come across this type of questions.
calculus limits functions
edited Jul 28 at 9:09
Javi
2,1481725
asked Jul 28 at 8:49
Bilal Sheikh
101
 |Â
show 2 more comments
up vote
0
down vote
favorite
I am learning limits these days and I have reached at finding limits by direct substitution but there's one that doesn't making sense to me at all. Let me illustrate it with an example $f(x) = x/x$ .The limit ( as $x$ approaches $0$) in this case does exist but by direct substitution, it says limit does not exist. What to do when you come across this type of questions.
calculus limits functions
edited Jul 28 at 9:09
Javi
2,1481725
asked Jul 28 at 8:49
Bilal Sheikh
101
1
You can cross out the $x$ and get $1$. This is a really broad question, but usually it is a substitution or an algebraic manipulation. Keep in mind that if the function is not continuous then you cannot use direct substitution.
– Sorfosh
Jul 28 at 8:52
3
Direct substitution? whatever that is it sounds worrying...
– Lord Shark the Unknown
Jul 28 at 9:08
You cannot use substitution in a point where the functions is not defined.
– Javi
Jul 28 at 9:09
Could you explain what the "direct substitution" thing you're talking about is? What do you substitute with what where, and how does that tell you that the limit doesn't exist? We can't tell you how your reasoning goes wrong unless you show us that reasoning.
– Henning Makholm
Jul 28 at 9:16
Let me rephrase my question, in which circumstances you can't find limit by direct substitution?
– Bilal Sheikh
Jul 28 at 9:24
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am learning limits these days and I have reached at finding limits by direct substitution but there's one that doesn't making sense to me at all. Let me illustrate it with an example $f(x) = x/x$ .The limit ( as $x$ approaches $0$) in this case does exist but by direct substitution, it says limit does not exist. What to do when you come across this type of questions.
calculus limits functions
edited Jul 28 at 9:09
Javi
2,1481725
asked Jul 28 at 8:49
Bilal Sheikh
101
I am learning limits these days and I have reached at finding limits by direct substitution but there's one that doesn't making sense to me at all. Let me illustrate it with an example $f(x) = x/x$ .The limit ( as $x$ approaches $0$) in this case does exist but by direct substitution, it says limit does not exist. What to do when you come across this type of questions.
calculus limits functions
edited Jul 28 at 9:09
Javi
2,1481725
asked Jul 28 at 8:49
Bilal Sheikh
101
edited Jul 28 at 9:09
Javi
2,1481725
edited Jul 28 at 9:09
Javi
2,1481725
edited Jul 28 at 9:09
Javi
2,1481725
2,1481725
asked Jul 28 at 8:49
Bilal Sheikh
101
asked Jul 28 at 8:49
Bilal Sheikh
101
asked Jul 28 at 8:49
Bilal Sheikh
101
101
1
You can cross out the $x$ and get $1$. This is a really broad question, but usually it is a substitution or an algebraic manipulation. Keep in mind that if the function is not continuous then you cannot use direct substitution.
– Sorfosh
Jul 28 at 8:52
3
Direct substitution? whatever that is it sounds worrying...
– Lord Shark the Unknown
Jul 28 at 9:08
You cannot use substitution in a point where the functions is not defined.
– Javi
Jul 28 at 9:09
Could you explain what the "direct substitution" thing you're talking about is? What do you substitute with what where, and how does that tell you that the limit doesn't exist? We can't tell you how your reasoning goes wrong unless you show us that reasoning.
– Henning Makholm
Jul 28 at 9:16
Let me rephrase my question, in which circumstances you can't find limit by direct substitution?
– Bilal Sheikh
Jul 28 at 9:24
 |Â
show 2 more comments
1
You can cross out the $x$ and get $1$. This is a really broad question, but usually it is a substitution or an algebraic manipulation. Keep in mind that if the function is not continuous then you cannot use direct substitution.
– Sorfosh
Jul 28 at 8:52
3
Direct substitution? whatever that is it sounds worrying...
– Lord Shark the Unknown
Jul 28 at 9:08
You cannot use substitution in a point where the functions is not defined.
– Javi
Jul 28 at 9:09
Could you explain what the "direct substitution" thing you're talking about is? What do you substitute with what where, and how does that tell you that the limit doesn't exist? We can't tell you how your reasoning goes wrong unless you show us that reasoning.
– Henning Makholm
Jul 28 at 9:16
Let me rephrase my question, in which circumstances you can't find limit by direct substitution?
– Bilal Sheikh
Jul 28 at 9:24
1
1
You can cross out the $x$ and get $1$. This is a really broad question, but usually it is a substitution or an algebraic manipulation. Keep in mind that if the function is not continuous then you cannot use direct substitution.
– Sorfosh
Jul 28 at 8:52
You can cross out the $x$ and get $1$. This is a really broad question, but usually it is a substitution or an algebraic manipulation. Keep in mind that if the function is not continuous then you cannot use direct substitution.
– Sorfosh
Jul 28 at 8:52
3
3
Direct substitution? whatever that is it sounds worrying...
– Lord Shark the Unknown
Jul 28 at 9:08
Direct substitution? whatever that is it sounds worrying...
– Lord Shark the Unknown
Jul 28 at 9:08
You cannot use substitution in a point where the functions is not defined.
– Javi
Jul 28 at 9:09
You cannot use substitution in a point where the functions is not defined.
– Javi
Jul 28 at 9:09
Could you explain what the "direct substitution" thing you're talking about is? What do you substitute with what where, and how does that tell you that the limit doesn't exist? We can't tell you how your reasoning goes wrong unless you show us that reasoning.
– Henning Makholm
Jul 28 at 9:16
Could you explain what the "direct substitution" thing you're talking about is? What do you substitute with what where, and how does that tell you that the limit doesn't exist? We can't tell you how your reasoning goes wrong unless you show us that reasoning.
– Henning Makholm
Jul 28 at 9:16
Let me rephrase my question, in which circumstances you can't find limit by direct substitution?
– Bilal Sheikh
Jul 28 at 9:24
Let me rephrase my question, in which circumstances you can't find limit by direct substitution?
– Bilal Sheikh
Jul 28 at 9:24
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
In the example you gave, as you're taking the limit when $;xto0;$ , it is your right and duty to consider number "very close" to zero but different from zero. Under this understanding, we simply have $;cfrac xx=1;$ for each number like this,and we can then easily deduce (according to its definition) the limit, which is 1.
In some other cases, as with $;cfracsin xx;$ , taking close values of $;x;$ to zero gives values close to $;1;$ , yet the formal proof that $;limlimits_xto0cfracsin xx=1;$ is a little messier, though much prettier, and it is usually done by geometric arguments.
Resuming: every case must be considered separatedly, but sometimes, as in your example, you can do simple arithmetic manipulations and get your result pretty simply.
answered Jul 28 at 9:10
DonAntonio
172k1483217
Let's take f(x) = x+2/x-1 Now, by direct substitution , you can't find limit (as x approaches 1). But, in real, it does exist
– Bilal Sheikh
Jul 28 at 9:39
@BilalSheikh The limit of your function at $1$ does not exist. The limit on the right ($x>1$) is $+infty$, whereas the limit on the left ($x<1$) is $-infty$.
– Suzet
Jul 28 at 9:46
add a comment |Â
up vote
0
down vote
Use other methods!
I don't hope anybody has tried to tell you that substitution works when the expression isn't defined, because (as your example shows) it doesn't.
In this case, you just need a simple algebraic manipulation to transform the expression you had to find the limit of into something else that is defined at the point (and therefore it's different), where you can "substitute" (there are no $x$'s left so it's a very trivial substitution in this case).
answered Jul 28 at 9:10
Henrik
5,63771930
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
In the example you gave, as you're taking the limit when $;xto0;$ , it is your right and duty to consider number "very close" to zero but different from zero. Under this understanding, we simply have $;cfrac xx=1;$ for each number like this,and we can then easily deduce (according to its definition) the limit, which is 1.
In some other cases, as with $;cfracsin xx;$ , taking close values of $;x;$ to zero gives values close to $;1;$ , yet the formal proof that $;limlimits_xto0cfracsin xx=1;$ is a little messier, though much prettier, and it is usually done by geometric arguments.
Resuming: every case must be considered separatedly, but sometimes, as in your example, you can do simple arithmetic manipulations and get your result pretty simply.
answered Jul 28 at 9:10
DonAntonio
172k1483217
Let's take f(x) = x+2/x-1 Now, by direct substitution , you can't find limit (as x approaches 1). But, in real, it does exist
– Bilal Sheikh
Jul 28 at 9:39
@BilalSheikh The limit of your function at $1$ does not exist. The limit on the right ($x>1$) is $+infty$, whereas the limit on the left ($x<1$) is $-infty$.
– Suzet
Jul 28 at 9:46
add a comment |Â
up vote
1
down vote
In the example you gave, as you're taking the limit when $;xto0;$ , it is your right and duty to consider number "very close" to zero but different from zero. Under this understanding, we simply have $;cfrac xx=1;$ for each number like this,and we can then easily deduce (according to its definition) the limit, which is 1.
In some other cases, as with $;cfracsin xx;$ , taking close values of $;x;$ to zero gives values close to $;1;$ , yet the formal proof that $;limlimits_xto0cfracsin xx=1;$ is a little messier, though much prettier, and it is usually done by geometric arguments.
Resuming: every case must be considered separatedly, but sometimes, as in your example, you can do simple arithmetic manipulations and get your result pretty simply.
answered Jul 28 at 9:10
DonAntonio
172k1483217
Let's take f(x) = x+2/x-1 Now, by direct substitution , you can't find limit (as x approaches 1). But, in real, it does exist
– Bilal Sheikh
Jul 28 at 9:39
@BilalSheikh The limit of your function at $1$ does not exist. The limit on the right ($x>1$) is $+infty$, whereas the limit on the left ($x<1$) is $-infty$.
– Suzet
Jul 28 at 9:46
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In the example you gave, as you're taking the limit when $;xto0;$ , it is your right and duty to consider number "very close" to zero but different from zero. Under this understanding, we simply have $;cfrac xx=1;$ for each number like this,and we can then easily deduce (according to its definition) the limit, which is 1.
In some other cases, as with $;cfracsin xx;$ , taking close values of $;x;$ to zero gives values close to $;1;$ , yet the formal proof that $;limlimits_xto0cfracsin xx=1;$ is a little messier, though much prettier, and it is usually done by geometric arguments.
Resuming: every case must be considered separatedly, but sometimes, as in your example, you can do simple arithmetic manipulations and get your result pretty simply.
answered Jul 28 at 9:10
DonAntonio
172k1483217
In the example you gave, as you're taking the limit when $;xto0;$ , it is your right and duty to consider number "very close" to zero but different from zero. Under this understanding, we simply have $;cfrac xx=1;$ for each number like this,and we can then easily deduce (according to its definition) the limit, which is 1.
In some other cases, as with $;cfracsin xx;$ , taking close values of $;x;$ to zero gives values close to $;1;$ , yet the formal proof that $;limlimits_xto0cfracsin xx=1;$ is a little messier, though much prettier, and it is usually done by geometric arguments.
Resuming: every case must be considered separatedly, but sometimes, as in your example, you can do simple arithmetic manipulations and get your result pretty simply.
answered Jul 28 at 9:10
DonAntonio
172k1483217
answered Jul 28 at 9:10
DonAntonio
172k1483217
answered Jul 28 at 9:10
DonAntonio
172k1483217
answered Jul 28 at 9:10
DonAntonio
172k1483217
172k1483217
Let's take f(x) = x+2/x-1 Now, by direct substitution , you can't find limit (as x approaches 1). But, in real, it does exist
– Bilal Sheikh
Jul 28 at 9:39
@BilalSheikh The limit of your function at $1$ does not exist. The limit on the right ($x>1$) is $+infty$, whereas the limit on the left ($x<1$) is $-infty$.
– Suzet
Jul 28 at 9:46
add a comment |Â
Let's take f(x) = x+2/x-1 Now, by direct substitution , you can't find limit (as x approaches 1). But, in real, it does exist
– Bilal Sheikh
Jul 28 at 9:39
@BilalSheikh The limit of your function at $1$ does not exist. The limit on the right ($x>1$) is $+infty$, whereas the limit on the left ($x<1$) is $-infty$.
– Suzet
Jul 28 at 9:46
Let's take f(x) = x+2/x-1 Now, by direct substitution , you can't find limit (as x approaches 1). But, in real, it does exist
– Bilal Sheikh
Jul 28 at 9:39
Let's take f(x) = x+2/x-1 Now, by direct substitution , you can't find limit (as x approaches 1). But, in real, it does exist
– Bilal Sheikh
Jul 28 at 9:39
@BilalSheikh The limit of your function at $1$ does not exist. The limit on the right ($x>1$) is $+infty$, whereas the limit on the left ($x<1$) is $-infty$.
– Suzet
Jul 28 at 9:46
@BilalSheikh The limit of your function at $1$ does not exist. The limit on the right ($x>1$) is $+infty$, whereas the limit on the left ($x<1$) is $-infty$.
– Suzet
Jul 28 at 9:46
add a comment |Â
up vote
0
down vote
Use other methods!
I don't hope anybody has tried to tell you that substitution works when the expression isn't defined, because (as your example shows) it doesn't.
In this case, you just need a simple algebraic manipulation to transform the expression you had to find the limit of into something else that is defined at the point (and therefore it's different), where you can "substitute" (there are no $x$'s left so it's a very trivial substitution in this case).
answered Jul 28 at 9:10
Henrik
5,63771930
add a comment |Â
up vote
0
down vote
Use other methods!
I don't hope anybody has tried to tell you that substitution works when the expression isn't defined, because (as your example shows) it doesn't.
In this case, you just need a simple algebraic manipulation to transform the expression you had to find the limit of into something else that is defined at the point (and therefore it's different), where you can "substitute" (there are no $x$'s left so it's a very trivial substitution in this case).
answered Jul 28 at 9:10
Henrik
5,63771930
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Use other methods!
I don't hope anybody has tried to tell you that substitution works when the expression isn't defined, because (as your example shows) it doesn't.
In this case, you just need a simple algebraic manipulation to transform the expression you had to find the limit of into something else that is defined at the point (and therefore it's different), where you can "substitute" (there are no $x$'s left so it's a very trivial substitution in this case).
answered Jul 28 at 9:10
Henrik
5,63771930
Use other methods!
I don't hope anybody has tried to tell you that substitution works when the expression isn't defined, because (as your example shows) it doesn't.
In this case, you just need a simple algebraic manipulation to transform the expression you had to find the limit of into something else that is defined at the point (and therefore it's different), where you can "substitute" (there are no $x$'s left so it's a very trivial substitution in this case).
answered Jul 28 at 9:10
Henrik
5,63771930
answered Jul 28 at 9:10
Henrik
5,63771930
answered Jul 28 at 9:10
Henrik
5,63771930
answered Jul 28 at 9:10
Henrik
5,63771930
5,63771930
add a comment |Â
add a comment |Â
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1
You can cross out the $x$ and get $1$. This is a really broad question, but usually it is a substitution or an algebraic manipulation. Keep in mind that if the function is not continuous then you cannot use direct substitution.
– Sorfosh
Jul 28 at 8:52
3
Direct substitution? whatever that is it sounds worrying...
– Lord Shark the Unknown
Jul 28 at 9:08
You cannot use substitution in a point where the functions is not defined.
– Javi
Jul 28 at 9:09
Could you explain what the "direct substitution" thing you're talking about is? What do you substitute with what where, and how does that tell you that the limit doesn't exist? We can't tell you how your reasoning goes wrong unless you show us that reasoning.
– Henning Makholm
Jul 28 at 9:16
Let me rephrase my question, in which circumstances you can't find limit by direct substitution?
– Bilal Sheikh
Jul 28 at 9:24