Find the value of $f(x)$ with proof.
$begingroup$
The equation is:
$$lim_{x to 2} frac{f(x) - 1}{x - 2} = 3$$
Suppose f(x) is a function. Find (with proof) the value of lim x→2 f(x), or conclude that this limit diverges.
Our professor wrote in a hint stating: Use Arithmetic of Limits and contradiction. Why can't $lim_{xto2}
f(x) = 0$, for
instance?
It is a question from a previous midterm and noone ive spoken to has solved it,so i wanna ask you guys.
real-analysis
$endgroup$
add a comment |
$begingroup$
The equation is:
$$lim_{x to 2} frac{f(x) - 1}{x - 2} = 3$$
Suppose f(x) is a function. Find (with proof) the value of lim x→2 f(x), or conclude that this limit diverges.
Our professor wrote in a hint stating: Use Arithmetic of Limits and contradiction. Why can't $lim_{xto2}
f(x) = 0$, for
instance?
It is a question from a previous midterm and noone ive spoken to has solved it,so i wanna ask you guys.
real-analysis
$endgroup$
$begingroup$
Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
$endgroup$
– DonAntonio
Jan 12 at 18:12
$begingroup$
Its a function on x.
$endgroup$
– user569685
Jan 12 at 18:16
$begingroup$
Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
$endgroup$
– DonAntonio
Jan 12 at 18:16
1
$begingroup$
@ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
$endgroup$
– DonAntonio
Jan 12 at 18:17
$begingroup$
Yes, you're right. I didn't read the question properly.@DonAntonio
$endgroup$
– Thomas Shelby
Jan 12 at 18:19
add a comment |
$begingroup$
The equation is:
$$lim_{x to 2} frac{f(x) - 1}{x - 2} = 3$$
Suppose f(x) is a function. Find (with proof) the value of lim x→2 f(x), or conclude that this limit diverges.
Our professor wrote in a hint stating: Use Arithmetic of Limits and contradiction. Why can't $lim_{xto2}
f(x) = 0$, for
instance?
It is a question from a previous midterm and noone ive spoken to has solved it,so i wanna ask you guys.
real-analysis
$endgroup$
The equation is:
$$lim_{x to 2} frac{f(x) - 1}{x - 2} = 3$$
Suppose f(x) is a function. Find (with proof) the value of lim x→2 f(x), or conclude that this limit diverges.
Our professor wrote in a hint stating: Use Arithmetic of Limits and contradiction. Why can't $lim_{xto2}
f(x) = 0$, for
instance?
It is a question from a previous midterm and noone ive spoken to has solved it,so i wanna ask you guys.
real-analysis
real-analysis
edited Jan 12 at 18:26
user569685
asked Jan 12 at 18:05
user569685user569685
102
102
$begingroup$
Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
$endgroup$
– DonAntonio
Jan 12 at 18:12
$begingroup$
Its a function on x.
$endgroup$
– user569685
Jan 12 at 18:16
$begingroup$
Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
$endgroup$
– DonAntonio
Jan 12 at 18:16
1
$begingroup$
@ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
$endgroup$
– DonAntonio
Jan 12 at 18:17
$begingroup$
Yes, you're right. I didn't read the question properly.@DonAntonio
$endgroup$
– Thomas Shelby
Jan 12 at 18:19
add a comment |
$begingroup$
Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
$endgroup$
– DonAntonio
Jan 12 at 18:12
$begingroup$
Its a function on x.
$endgroup$
– user569685
Jan 12 at 18:16
$begingroup$
Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
$endgroup$
– DonAntonio
Jan 12 at 18:16
1
$begingroup$
@ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
$endgroup$
– DonAntonio
Jan 12 at 18:17
$begingroup$
Yes, you're right. I didn't read the question properly.@DonAntonio
$endgroup$
– Thomas Shelby
Jan 12 at 18:19
$begingroup$
Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
$endgroup$
– DonAntonio
Jan 12 at 18:12
$begingroup$
Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
$endgroup$
– DonAntonio
Jan 12 at 18:12
$begingroup$
Its a function on x.
$endgroup$
– user569685
Jan 12 at 18:16
$begingroup$
Its a function on x.
$endgroup$
– user569685
Jan 12 at 18:16
$begingroup$
Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
$endgroup$
– DonAntonio
Jan 12 at 18:16
$begingroup$
Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
$endgroup$
– DonAntonio
Jan 12 at 18:16
1
1
$begingroup$
@ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
$endgroup$
– DonAntonio
Jan 12 at 18:17
$begingroup$
@ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
$endgroup$
– DonAntonio
Jan 12 at 18:17
$begingroup$
Yes, you're right. I didn't read the question properly.@DonAntonio
$endgroup$
– Thomas Shelby
Jan 12 at 18:19
$begingroup$
Yes, you're right. I didn't read the question properly.@DonAntonio
$endgroup$
– Thomas Shelby
Jan 12 at 18:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint
We know that
$$
lim_{xto 2}(f(x)-1)=left(lim_{xto 2}frac{f(x)-1}{x-2}right)timeslim_{xto2}(x-2)tag{0}
$$
by the product rule for limits and
$$
lim_{xto2}f(x)=lim_{xto 2}(f(x)-1)+lim_{xto2}1tag{1}.
$$
You have enough information to compute all the limits.
$endgroup$
$begingroup$
Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
$endgroup$
– DonAntonio
Jan 12 at 18:15
$begingroup$
@user569685 Great! Why then in the world didn't you write this in your question?!
$endgroup$
– DonAntonio
Jan 12 at 18:25
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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$begingroup$
Hint
We know that
$$
lim_{xto 2}(f(x)-1)=left(lim_{xto 2}frac{f(x)-1}{x-2}right)timeslim_{xto2}(x-2)tag{0}
$$
by the product rule for limits and
$$
lim_{xto2}f(x)=lim_{xto 2}(f(x)-1)+lim_{xto2}1tag{1}.
$$
You have enough information to compute all the limits.
$endgroup$
$begingroup$
Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
$endgroup$
– DonAntonio
Jan 12 at 18:15
$begingroup$
@user569685 Great! Why then in the world didn't you write this in your question?!
$endgroup$
– DonAntonio
Jan 12 at 18:25
add a comment |
$begingroup$
Hint
We know that
$$
lim_{xto 2}(f(x)-1)=left(lim_{xto 2}frac{f(x)-1}{x-2}right)timeslim_{xto2}(x-2)tag{0}
$$
by the product rule for limits and
$$
lim_{xto2}f(x)=lim_{xto 2}(f(x)-1)+lim_{xto2}1tag{1}.
$$
You have enough information to compute all the limits.
$endgroup$
$begingroup$
Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
$endgroup$
– DonAntonio
Jan 12 at 18:15
$begingroup$
@user569685 Great! Why then in the world didn't you write this in your question?!
$endgroup$
– DonAntonio
Jan 12 at 18:25
add a comment |
$begingroup$
Hint
We know that
$$
lim_{xto 2}(f(x)-1)=left(lim_{xto 2}frac{f(x)-1}{x-2}right)timeslim_{xto2}(x-2)tag{0}
$$
by the product rule for limits and
$$
lim_{xto2}f(x)=lim_{xto 2}(f(x)-1)+lim_{xto2}1tag{1}.
$$
You have enough information to compute all the limits.
$endgroup$
Hint
We know that
$$
lim_{xto 2}(f(x)-1)=left(lim_{xto 2}frac{f(x)-1}{x-2}right)timeslim_{xto2}(x-2)tag{0}
$$
by the product rule for limits and
$$
lim_{xto2}f(x)=lim_{xto 2}(f(x)-1)+lim_{xto2}1tag{1}.
$$
You have enough information to compute all the limits.
answered Jan 12 at 18:08
Foobaz JohnFoobaz John
21.9k41352
21.9k41352
$begingroup$
Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
$endgroup$
– DonAntonio
Jan 12 at 18:15
$begingroup$
@user569685 Great! Why then in the world didn't you write this in your question?!
$endgroup$
– DonAntonio
Jan 12 at 18:25
add a comment |
$begingroup$
Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
$endgroup$
– DonAntonio
Jan 12 at 18:15
$begingroup$
@user569685 Great! Why then in the world didn't you write this in your question?!
$endgroup$
– DonAntonio
Jan 12 at 18:25
$begingroup$
Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
$endgroup$
– DonAntonio
Jan 12 at 18:15
$begingroup$
Yes, the limits can be computed, apparently (if the OP did give all the relevant info...), yet the question seems to be what is the value of $;f;$ ...perhaps when $;x=2;$ or something like that.
$endgroup$
– DonAntonio
Jan 12 at 18:15
$begingroup$
@user569685 Great! Why then in the world didn't you write this in your question?!
$endgroup$
– DonAntonio
Jan 12 at 18:25
$begingroup$
@user569685 Great! Why then in the world didn't you write this in your question?!
$endgroup$
– DonAntonio
Jan 12 at 18:25
add a comment |
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$begingroup$
Didn't you forget a "little" detail? What is $;f(x);$ , anyway?!
$endgroup$
– DonAntonio
Jan 12 at 18:12
$begingroup$
Its a function on x.
$endgroup$
– user569685
Jan 12 at 18:16
$begingroup$
Then perhaps the actual question is "what is the value of $;f(2);$ ..."?? Because the value of $;f(x);$ cannot be known from the given info.
$endgroup$
– DonAntonio
Jan 12 at 18:16
1
$begingroup$
@ThomasShelby I saw it...read my comment below such answer. Apparently the OP meant to ask what is the value of $;f(2);$ ...
$endgroup$
– DonAntonio
Jan 12 at 18:17
$begingroup$
Yes, you're right. I didn't read the question properly.@DonAntonio
$endgroup$
– Thomas Shelby
Jan 12 at 18:19