Confused about how to find horizontal asymptote when sketching a curve
I was taught in high school that if the numerator's degree was higher than the denominator's, then there was no H.A, if they were the same it was the coefficient's over coefficient, and if the denominator's degree was higher it was $y = 0$. But now in first year uni I'm being taught that it's when the limit as x approaches negative and positive infinity, and I guess that makes sense, it's just testing if it's bounded above or below by anything, but I just wanna clarify some things. For example $e^x$. As x approaches negative infinity, the limit is 0. As x approaches positive infinity the limit does not exist. So if my teacher just says the H.A is where the limit is at x approaches positive AND negative do I just state both? Or the one that has the limit? What am I suppose to say? What about for $f(x) = sin(x)$ where it is bounded above and below, do I just say the H.A is at $y = 1$ and $y = -1$?
calculus
asked Nov 29 at 16:54
ming
3165
add a comment |
I was taught in high school that if the numerator's degree was higher than the denominator's, then there was no H.A, if they were the same it was the coefficient's over coefficient, and if the denominator's degree was higher it was $y = 0$. But now in first year uni I'm being taught that it's when the limit as x approaches negative and positive infinity, and I guess that makes sense, it's just testing if it's bounded above or below by anything, but I just wanna clarify some things. For example $e^x$. As x approaches negative infinity, the limit is 0. As x approaches positive infinity the limit does not exist. So if my teacher just says the H.A is where the limit is at x approaches positive AND negative do I just state both? Or the one that has the limit? What am I suppose to say? What about for $f(x) = sin(x)$ where it is bounded above and below, do I just say the H.A is at $y = 1$ and $y = -1$?
calculus
asked Nov 29 at 16:54
ming
3165
add a comment |
I was taught in high school that if the numerator's degree was higher than the denominator's, then there was no H.A, if they were the same it was the coefficient's over coefficient, and if the denominator's degree was higher it was $y = 0$. But now in first year uni I'm being taught that it's when the limit as x approaches negative and positive infinity, and I guess that makes sense, it's just testing if it's bounded above or below by anything, but I just wanna clarify some things. For example $e^x$. As x approaches negative infinity, the limit is 0. As x approaches positive infinity the limit does not exist. So if my teacher just says the H.A is where the limit is at x approaches positive AND negative do I just state both? Or the one that has the limit? What am I suppose to say? What about for $f(x) = sin(x)$ where it is bounded above and below, do I just say the H.A is at $y = 1$ and $y = -1$?
calculus
asked Nov 29 at 16:54
ming
3165
I was taught in high school that if the numerator's degree was higher than the denominator's, then there was no H.A, if they were the same it was the coefficient's over coefficient, and if the denominator's degree was higher it was $y = 0$. But now in first year uni I'm being taught that it's when the limit as x approaches negative and positive infinity, and I guess that makes sense, it's just testing if it's bounded above or below by anything, but I just wanna clarify some things. For example $e^x$. As x approaches negative infinity, the limit is 0. As x approaches positive infinity the limit does not exist. So if my teacher just says the H.A is where the limit is at x approaches positive AND negative do I just state both? Or the one that has the limit? What am I suppose to say? What about for $f(x) = sin(x)$ where it is bounded above and below, do I just say the H.A is at $y = 1$ and $y = -1$?
calculus
calculus
asked Nov 29 at 16:54
ming
3165
asked Nov 29 at 16:54
ming
3165
asked Nov 29 at 16:54
ming
3165
asked Nov 29 at 16:54
ming
3165
asked Nov 29 at 16:54
ming
3165
3165
add a comment |
add a comment |
1 Answer
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It sounds like the rule you were taught in high school was for functions defined as one polynomial divided by another - also known as "rational functions". Now you are dealing with more general types of functions, so your high school rule will not always apply.
A more general definition of a horizontal asymptote is that it is a limiting value that $f(x)$ approaches as $x rightarrow infty$ or as $x rightarrow -infty$.
So $f(x)=e^x$ has a horizontal asymptote $f(x) rightarrow 0$ as $x rightarrow -infty$, but no horiozontal asymptote as $x rightarrow infty$, whereas $f(x)= frac{1}{x}$ has a horizontal asymptote $f(x) rightarrow 0$ as $x rightarrow infty$ and as $x rightarrow -infty$.
Sometimes the asymptotes at $infty$ and $-infty$ may have different values. For example, if $f(x) = tan^{-1}(x)$ then $f(x) rightarrow -frac{pi}{2}$ as $x rightarrow -infty$, and $f(x) rightarrow frac{pi}{2}$ as $x rightarrow infty$.
However, $f(x) = sin(x)$ has no horizontal asymptotes. $sin(x)$ oscillates between $-1$ and $+1$ but does not stay close to either value. These informal notions will become more precise when you learn about the formal definition of limits.
answered Nov 29 at 17:14
gandalf61
7,663623
So when I need to draw the curve of $f(x) = e^x$ do I officially state "Horizontal Asymptote at x = 0?"
– ming
Nov 29 at 18:54
And vertical asymptotes are just basically the restrictions stated in the domain right?
– ming
Nov 29 at 19:00
@ming horizontal asymptote is considered only when $xtopminfty$
– user376343
Nov 29 at 22:48
add a comment |
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It sounds like the rule you were taught in high school was for functions defined as one polynomial divided by another - also known as "rational functions". Now you are dealing with more general types of functions, so your high school rule will not always apply.
A more general definition of a horizontal asymptote is that it is a limiting value that $f(x)$ approaches as $x rightarrow infty$ or as $x rightarrow -infty$.
So $f(x)=e^x$ has a horizontal asymptote $f(x) rightarrow 0$ as $x rightarrow -infty$, but no horiozontal asymptote as $x rightarrow infty$, whereas $f(x)= frac{1}{x}$ has a horizontal asymptote $f(x) rightarrow 0$ as $x rightarrow infty$ and as $x rightarrow -infty$.
Sometimes the asymptotes at $infty$ and $-infty$ may have different values. For example, if $f(x) = tan^{-1}(x)$ then $f(x) rightarrow -frac{pi}{2}$ as $x rightarrow -infty$, and $f(x) rightarrow frac{pi}{2}$ as $x rightarrow infty$.
However, $f(x) = sin(x)$ has no horizontal asymptotes. $sin(x)$ oscillates between $-1$ and $+1$ but does not stay close to either value. These informal notions will become more precise when you learn about the formal definition of limits.
answered Nov 29 at 17:14
gandalf61
7,663623
So when I need to draw the curve of $f(x) = e^x$ do I officially state "Horizontal Asymptote at x = 0?"
– ming
Nov 29 at 18:54
And vertical asymptotes are just basically the restrictions stated in the domain right?
– ming
Nov 29 at 19:00
@ming horizontal asymptote is considered only when $xtopminfty$
– user376343
Nov 29 at 22:48
add a comment |
It sounds like the rule you were taught in high school was for functions defined as one polynomial divided by another - also known as "rational functions". Now you are dealing with more general types of functions, so your high school rule will not always apply.
A more general definition of a horizontal asymptote is that it is a limiting value that $f(x)$ approaches as $x rightarrow infty$ or as $x rightarrow -infty$.
So $f(x)=e^x$ has a horizontal asymptote $f(x) rightarrow 0$ as $x rightarrow -infty$, but no horiozontal asymptote as $x rightarrow infty$, whereas $f(x)= frac{1}{x}$ has a horizontal asymptote $f(x) rightarrow 0$ as $x rightarrow infty$ and as $x rightarrow -infty$.
Sometimes the asymptotes at $infty$ and $-infty$ may have different values. For example, if $f(x) = tan^{-1}(x)$ then $f(x) rightarrow -frac{pi}{2}$ as $x rightarrow -infty$, and $f(x) rightarrow frac{pi}{2}$ as $x rightarrow infty$.
However, $f(x) = sin(x)$ has no horizontal asymptotes. $sin(x)$ oscillates between $-1$ and $+1$ but does not stay close to either value. These informal notions will become more precise when you learn about the formal definition of limits.
answered Nov 29 at 17:14
gandalf61
7,663623
So when I need to draw the curve of $f(x) = e^x$ do I officially state "Horizontal Asymptote at x = 0?"
– ming
Nov 29 at 18:54
And vertical asymptotes are just basically the restrictions stated in the domain right?
– ming
Nov 29 at 19:00
@ming horizontal asymptote is considered only when $xtopminfty$
– user376343
Nov 29 at 22:48
add a comment |
It sounds like the rule you were taught in high school was for functions defined as one polynomial divided by another - also known as "rational functions". Now you are dealing with more general types of functions, so your high school rule will not always apply.
A more general definition of a horizontal asymptote is that it is a limiting value that $f(x)$ approaches as $x rightarrow infty$ or as $x rightarrow -infty$.
So $f(x)=e^x$ has a horizontal asymptote $f(x) rightarrow 0$ as $x rightarrow -infty$, but no horiozontal asymptote as $x rightarrow infty$, whereas $f(x)= frac{1}{x}$ has a horizontal asymptote $f(x) rightarrow 0$ as $x rightarrow infty$ and as $x rightarrow -infty$.
Sometimes the asymptotes at $infty$ and $-infty$ may have different values. For example, if $f(x) = tan^{-1}(x)$ then $f(x) rightarrow -frac{pi}{2}$ as $x rightarrow -infty$, and $f(x) rightarrow frac{pi}{2}$ as $x rightarrow infty$.
However, $f(x) = sin(x)$ has no horizontal asymptotes. $sin(x)$ oscillates between $-1$ and $+1$ but does not stay close to either value. These informal notions will become more precise when you learn about the formal definition of limits.
answered Nov 29 at 17:14
gandalf61
7,663623
It sounds like the rule you were taught in high school was for functions defined as one polynomial divided by another - also known as "rational functions". Now you are dealing with more general types of functions, so your high school rule will not always apply.
A more general definition of a horizontal asymptote is that it is a limiting value that $f(x)$ approaches as $x rightarrow infty$ or as $x rightarrow -infty$.
So $f(x)=e^x$ has a horizontal asymptote $f(x) rightarrow 0$ as $x rightarrow -infty$, but no horiozontal asymptote as $x rightarrow infty$, whereas $f(x)= frac{1}{x}$ has a horizontal asymptote $f(x) rightarrow 0$ as $x rightarrow infty$ and as $x rightarrow -infty$.
Sometimes the asymptotes at $infty$ and $-infty$ may have different values. For example, if $f(x) = tan^{-1}(x)$ then $f(x) rightarrow -frac{pi}{2}$ as $x rightarrow -infty$, and $f(x) rightarrow frac{pi}{2}$ as $x rightarrow infty$.
However, $f(x) = sin(x)$ has no horizontal asymptotes. $sin(x)$ oscillates between $-1$ and $+1$ but does not stay close to either value. These informal notions will become more precise when you learn about the formal definition of limits.
answered Nov 29 at 17:14
gandalf61
7,663623
edited Nov 29 at 17:38
answered Nov 29 at 17:14
gandalf61
7,663623
answered Nov 29 at 17:14
gandalf61
7,663623
answered Nov 29 at 17:14
gandalf61
7,663623
7,663623
So when I need to draw the curve of $f(x) = e^x$ do I officially state "Horizontal Asymptote at x = 0?"
– ming
Nov 29 at 18:54
And vertical asymptotes are just basically the restrictions stated in the domain right?
– ming
Nov 29 at 19:00
@ming horizontal asymptote is considered only when $xtopminfty$
– user376343
Nov 29 at 22:48
add a comment |
So when I need to draw the curve of $f(x) = e^x$ do I officially state "Horizontal Asymptote at x = 0?"
– ming
Nov 29 at 18:54
And vertical asymptotes are just basically the restrictions stated in the domain right?
– ming
Nov 29 at 19:00
@ming horizontal asymptote is considered only when $xtopminfty$
– user376343
Nov 29 at 22:48
So when I need to draw the curve of $f(x) = e^x$ do I officially state "Horizontal Asymptote at x = 0?"
– ming
Nov 29 at 18:54
So when I need to draw the curve of $f(x) = e^x$ do I officially state "Horizontal Asymptote at x = 0?"
– ming
Nov 29 at 18:54
And vertical asymptotes are just basically the restrictions stated in the domain right?
– ming
Nov 29 at 19:00
And vertical asymptotes are just basically the restrictions stated in the domain right?
– ming
Nov 29 at 19:00
@ming horizontal asymptote is considered only when $xtopminfty$
– user376343
Nov 29 at 22:48
@ming horizontal asymptote is considered only when $xtopminfty$
– user376343
Nov 29 at 22:48
add a comment |
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