Compute the limit $lim_{xto1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}$
$begingroup$
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}$$
What I tried: divided by $x^3$
$$lim_{xto1}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}$$
Then I plug in $x=1$
$$frac{2-2+1-1}{1-1+3-3}=frac{0}{0}$$
This is not correct, where did I make my mistake?
calculus limits rational-functions
edited Dec 8 '18 at 14:26
Rodrigo de Azevedo
12.9k41857
asked Dec 8 '18 at 14:20
Bili DebiliBili Debili
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$endgroup$
add a comment |
$begingroup$
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}$$
What I tried: divided by $x^3$
$$lim_{xto1}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}$$
Then I plug in $x=1$
$$frac{2-2+1-1}{1-1+3-3}=frac{0}{0}$$
This is not correct, where did I make my mistake?
calculus limits rational-functions
edited Dec 8 '18 at 14:26
Rodrigo de Azevedo
12.9k41857
asked Dec 8 '18 at 14:20
Bili DebiliBili Debili
1428
$endgroup$
add a comment |
$begingroup$
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}$$
What I tried: divided by $x^3$
$$lim_{xto1}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}$$
Then I plug in $x=1$
$$frac{2-2+1-1}{1-1+3-3}=frac{0}{0}$$
This is not correct, where did I make my mistake?
calculus limits rational-functions
edited Dec 8 '18 at 14:26
Rodrigo de Azevedo
12.9k41857
asked Dec 8 '18 at 14:20
Bili DebiliBili Debili
1428
$endgroup$
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}$$
What I tried: divided by $x^3$
$$lim_{xto1}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}$$
Then I plug in $x=1$
$$frac{2-2+1-1}{1-1+3-3}=frac{0}{0}$$
This is not correct, where did I make my mistake?
calculus limits rational-functions
calculus limits rational-functions
edited Dec 8 '18 at 14:26
Rodrigo de Azevedo
12.9k41857
asked Dec 8 '18 at 14:20
Bili DebiliBili Debili
1428
edited Dec 8 '18 at 14:26
Rodrigo de Azevedo
12.9k41857
asked Dec 8 '18 at 14:20
Bili DebiliBili Debili
1428
edited Dec 8 '18 at 14:26
Rodrigo de Azevedo
12.9k41857
edited Dec 8 '18 at 14:26
Rodrigo de Azevedo
12.9k41857
edited Dec 8 '18 at 14:26
Rodrigo de Azevedo
12.9k41857
12.9k41857
asked Dec 8 '18 at 14:20
Bili DebiliBili Debili
1428
asked Dec 8 '18 at 14:20
Bili DebiliBili Debili
1428
asked Dec 8 '18 at 14:20
Bili DebiliBili Debili
1428
1428
add a comment |
add a comment |
3 Answers
3
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oldest
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$begingroup$
We only evaluate it directly if we do not end up with indeterminate form.
Hint:
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}= lim_{xto 1}{frac{2x^2(x-1)+(x-1)}{x^2(x-1)+3(x-1)}}$$
answered Dec 8 '18 at 14:24
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
Since it looks like $0/0$, it means that both the denominator and numerator are zero. That means that $1$ is a root of both. What that tells you is that you should take $(x-1)$ as a factor, then simplify the fraction.
answered Dec 8 '18 at 14:26
AndreiAndrei
11.8k21026
$endgroup$
add a comment |
$begingroup$
That's not an effective way to solve since the limit is at $x=1$ and not to $infty$, indeed in that latter case your method would be fine since we obtain
$$lim_{xtoinfty}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}=2$$
As an effective alternative, in this case, we could use l'Hopital for example.
Edit for more detail requested
By l'Hopital we obtain
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}stackrel{H.R.}=lim_{xto 1}{frac{6x^2-4x+1}{3x^2-2x+3}}=ldots$$
Can you conclude?
answered Dec 8 '18 at 14:25
gimusigimusi
92.8k84494
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We only evaluate it directly if we do not end up with indeterminate form.
Hint:
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}= lim_{xto 1}{frac{2x^2(x-1)+(x-1)}{x^2(x-1)+3(x-1)}}$$
answered Dec 8 '18 at 14:24
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
We only evaluate it directly if we do not end up with indeterminate form.
Hint:
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}= lim_{xto 1}{frac{2x^2(x-1)+(x-1)}{x^2(x-1)+3(x-1)}}$$
answered Dec 8 '18 at 14:24
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
We only evaluate it directly if we do not end up with indeterminate form.
Hint:
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}= lim_{xto 1}{frac{2x^2(x-1)+(x-1)}{x^2(x-1)+3(x-1)}}$$
answered Dec 8 '18 at 14:24
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
We only evaluate it directly if we do not end up with indeterminate form.
Hint:
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}= lim_{xto 1}{frac{2x^2(x-1)+(x-1)}{x^2(x-1)+3(x-1)}}$$
answered Dec 8 '18 at 14:24
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 8 '18 at 14:24
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 8 '18 at 14:24
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 8 '18 at 14:24
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
add a comment |
add a comment |
$begingroup$
Since it looks like $0/0$, it means that both the denominator and numerator are zero. That means that $1$ is a root of both. What that tells you is that you should take $(x-1)$ as a factor, then simplify the fraction.
answered Dec 8 '18 at 14:26
AndreiAndrei
11.8k21026
$endgroup$
add a comment |
$begingroup$
Since it looks like $0/0$, it means that both the denominator and numerator are zero. That means that $1$ is a root of both. What that tells you is that you should take $(x-1)$ as a factor, then simplify the fraction.
answered Dec 8 '18 at 14:26
AndreiAndrei
11.8k21026
$endgroup$
add a comment |
$begingroup$
Since it looks like $0/0$, it means that both the denominator and numerator are zero. That means that $1$ is a root of both. What that tells you is that you should take $(x-1)$ as a factor, then simplify the fraction.
answered Dec 8 '18 at 14:26
AndreiAndrei
11.8k21026
$endgroup$
Since it looks like $0/0$, it means that both the denominator and numerator are zero. That means that $1$ is a root of both. What that tells you is that you should take $(x-1)$ as a factor, then simplify the fraction.
answered Dec 8 '18 at 14:26
AndreiAndrei
11.8k21026
answered Dec 8 '18 at 14:26
AndreiAndrei
11.8k21026
answered Dec 8 '18 at 14:26
AndreiAndrei
11.8k21026
answered Dec 8 '18 at 14:26
AndreiAndrei
11.8k21026
11.8k21026
add a comment |
add a comment |
$begingroup$
That's not an effective way to solve since the limit is at $x=1$ and not to $infty$, indeed in that latter case your method would be fine since we obtain
$$lim_{xtoinfty}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}=2$$
As an effective alternative, in this case, we could use l'Hopital for example.
Edit for more detail requested
By l'Hopital we obtain
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}stackrel{H.R.}=lim_{xto 1}{frac{6x^2-4x+1}{3x^2-2x+3}}=ldots$$
Can you conclude?
answered Dec 8 '18 at 14:25
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
That's not an effective way to solve since the limit is at $x=1$ and not to $infty$, indeed in that latter case your method would be fine since we obtain
$$lim_{xtoinfty}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}=2$$
As an effective alternative, in this case, we could use l'Hopital for example.
Edit for more detail requested
By l'Hopital we obtain
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}stackrel{H.R.}=lim_{xto 1}{frac{6x^2-4x+1}{3x^2-2x+3}}=ldots$$
Can you conclude?
answered Dec 8 '18 at 14:25
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
That's not an effective way to solve since the limit is at $x=1$ and not to $infty$, indeed in that latter case your method would be fine since we obtain
$$lim_{xtoinfty}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}=2$$
As an effective alternative, in this case, we could use l'Hopital for example.
Edit for more detail requested
By l'Hopital we obtain
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}stackrel{H.R.}=lim_{xto 1}{frac{6x^2-4x+1}{3x^2-2x+3}}=ldots$$
Can you conclude?
answered Dec 8 '18 at 14:25
gimusigimusi
92.8k84494
$endgroup$
That's not an effective way to solve since the limit is at $x=1$ and not to $infty$, indeed in that latter case your method would be fine since we obtain
$$lim_{xtoinfty}{frac{2-frac{2}{x}+frac{1}{x^2}-frac{1}{x^3}}{1-frac{1}{x}+frac{3}{x^2}-frac{3}{x^3}}}=2$$
As an effective alternative, in this case, we could use l'Hopital for example.
Edit for more detail requested
By l'Hopital we obtain
$$lim_{xto 1}{frac{2x^3-2x^2+x-1}{x^3-x^2+3x-3}}stackrel{H.R.}=lim_{xto 1}{frac{6x^2-4x+1}{3x^2-2x+3}}=ldots$$
Can you conclude?
answered Dec 8 '18 at 14:25
gimusigimusi
92.8k84494
edited Dec 8 '18 at 14:46
answered Dec 8 '18 at 14:25
gimusigimusi
92.8k84494
answered Dec 8 '18 at 14:25
gimusigimusi
92.8k84494
answered Dec 8 '18 at 14:25
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
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