Determining Slant Asymptote using $lim_{xto infty}frac{f(x)}{x}$ for $(1-x^3)^{1/3}$
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I have the following function for which I am able to determine the slant asymptote of by find the limit of this function over $x$ as $x to infty$,
$$(1-x^3)^{1/3}$$
The result of this limit is $-1$, so I know that is the coefficient, k, of the asymptote $y=kx+b$. Now I am trying to find the $b$ term by evaluating
$$lim_{xto infty}f(x) - kx = lim_{xto infty} (1-x^3)^{1/3} + x$$
I multiply by the conjugate and simplify as follows,
$$lim_{xtoinfty}frac{1-x^3}{x[(x^{-3}-1)^{1/3}-1]}$$
Now I divide out the X's and substitute the limit, but no matter what I get inifinity as the result of this limit. Does this mean the asymptote is simply $-x$ or am I doing something wrong?
calculus limits
asked Dec 7 '18 at 3:23
DanielleDanielle
2179
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add a comment |
$begingroup$
I have the following function for which I am able to determine the slant asymptote of by find the limit of this function over $x$ as $x to infty$,
$$(1-x^3)^{1/3}$$
The result of this limit is $-1$, so I know that is the coefficient, k, of the asymptote $y=kx+b$. Now I am trying to find the $b$ term by evaluating
$$lim_{xto infty}f(x) - kx = lim_{xto infty} (1-x^3)^{1/3} + x$$
I multiply by the conjugate and simplify as follows,
$$lim_{xtoinfty}frac{1-x^3}{x[(x^{-3}-1)^{1/3}-1]}$$
Now I divide out the X's and substitute the limit, but no matter what I get inifinity as the result of this limit. Does this mean the asymptote is simply $-x$ or am I doing something wrong?
calculus limits
asked Dec 7 '18 at 3:23
DanielleDanielle
2179
$endgroup$
2
$begingroup$
$$b=lim_{xto infty} ((1-x^3)^{1/3} + x) times dfrac{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{x^2left(sqrt[3]{(frac{1}{x^3}-1)^2} - sqrt[3]{frac{1}{x^3}-1}+1right)}=0$$
$endgroup$
– Nosrati
Dec 7 '18 at 4:00
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@Danielle your solution is right until the multiplication by the conjugate. You did it as if there were a square root, not a third. Nosrati wrote the correct one.
$endgroup$
– user376343
Dec 7 '18 at 11:43
add a comment |
$begingroup$
I have the following function for which I am able to determine the slant asymptote of by find the limit of this function over $x$ as $x to infty$,
$$(1-x^3)^{1/3}$$
The result of this limit is $-1$, so I know that is the coefficient, k, of the asymptote $y=kx+b$. Now I am trying to find the $b$ term by evaluating
$$lim_{xto infty}f(x) - kx = lim_{xto infty} (1-x^3)^{1/3} + x$$
I multiply by the conjugate and simplify as follows,
$$lim_{xtoinfty}frac{1-x^3}{x[(x^{-3}-1)^{1/3}-1]}$$
Now I divide out the X's and substitute the limit, but no matter what I get inifinity as the result of this limit. Does this mean the asymptote is simply $-x$ or am I doing something wrong?
calculus limits
asked Dec 7 '18 at 3:23
DanielleDanielle
2179
$endgroup$
I have the following function for which I am able to determine the slant asymptote of by find the limit of this function over $x$ as $x to infty$,
$$(1-x^3)^{1/3}$$
The result of this limit is $-1$, so I know that is the coefficient, k, of the asymptote $y=kx+b$. Now I am trying to find the $b$ term by evaluating
$$lim_{xto infty}f(x) - kx = lim_{xto infty} (1-x^3)^{1/3} + x$$
I multiply by the conjugate and simplify as follows,
$$lim_{xtoinfty}frac{1-x^3}{x[(x^{-3}-1)^{1/3}-1]}$$
Now I divide out the X's and substitute the limit, but no matter what I get inifinity as the result of this limit. Does this mean the asymptote is simply $-x$ or am I doing something wrong?
calculus limits
calculus limits
asked Dec 7 '18 at 3:23
DanielleDanielle
2179
asked Dec 7 '18 at 3:23
DanielleDanielle
2179
asked Dec 7 '18 at 3:23
DanielleDanielle
2179
asked Dec 7 '18 at 3:23
DanielleDanielle
2179
asked Dec 7 '18 at 3:23
DanielleDanielle
2179
2179
2
$begingroup$
$$b=lim_{xto infty} ((1-x^3)^{1/3} + x) times dfrac{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{x^2left(sqrt[3]{(frac{1}{x^3}-1)^2} - sqrt[3]{frac{1}{x^3}-1}+1right)}=0$$
$endgroup$
– Nosrati
Dec 7 '18 at 4:00
$begingroup$
@Danielle your solution is right until the multiplication by the conjugate. You did it as if there were a square root, not a third. Nosrati wrote the correct one.
$endgroup$
– user376343
Dec 7 '18 at 11:43
add a comment |
2
$begingroup$
$$b=lim_{xto infty} ((1-x^3)^{1/3} + x) times dfrac{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{x^2left(sqrt[3]{(frac{1}{x^3}-1)^2} - sqrt[3]{frac{1}{x^3}-1}+1right)}=0$$
$endgroup$
– Nosrati
Dec 7 '18 at 4:00
$begingroup$
@Danielle your solution is right until the multiplication by the conjugate. You did it as if there were a square root, not a third. Nosrati wrote the correct one.
$endgroup$
– user376343
Dec 7 '18 at 11:43
2
2
$begingroup$
$$b=lim_{xto infty} ((1-x^3)^{1/3} + x) times dfrac{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{x^2left(sqrt[3]{(frac{1}{x^3}-1)^2} - sqrt[3]{frac{1}{x^3}-1}+1right)}=0$$
$endgroup$
– Nosrati
Dec 7 '18 at 4:00
$begingroup$
$$b=lim_{xto infty} ((1-x^3)^{1/3} + x) times dfrac{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{x^2left(sqrt[3]{(frac{1}{x^3}-1)^2} - sqrt[3]{frac{1}{x^3}-1}+1right)}=0$$
$endgroup$
– Nosrati
Dec 7 '18 at 4:00
$begingroup$
@Danielle your solution is right until the multiplication by the conjugate. You did it as if there were a square root, not a third. Nosrati wrote the correct one.
$endgroup$
– user376343
Dec 7 '18 at 11:43
$begingroup$
@Danielle your solution is right until the multiplication by the conjugate. You did it as if there were a square root, not a third. Nosrati wrote the correct one.
$endgroup$
– user376343
Dec 7 '18 at 11:43
add a comment |
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$begingroup$
$$b=lim_{xto infty} ((1-x^3)^{1/3} + x) times dfrac{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{sqrt[3]{(1-x^3)^2} - xsqrt[3]{1-x^3}+x^2}=lim_{xto infty} dfrac{1}{x^2left(sqrt[3]{(frac{1}{x^3}-1)^2} - sqrt[3]{frac{1}{x^3}-1}+1right)}=0$$
$endgroup$
– Nosrati
Dec 7 '18 at 4:00
$begingroup$
@Danielle your solution is right until the multiplication by the conjugate. You did it as if there were a square root, not a third. Nosrati wrote the correct one.
$endgroup$
– user376343
Dec 7 '18 at 11:43