
How do I find the limit as $ x $ approaches negative infinity of a polynomial?
Answer
542.4k+ views
Hint: If the polynomial (a non-zero, real-valued function of the form $ c{x^n} $ - where $ c \in \mathbb{R}\backslash \{ 0\} $ and $ n \in \mathbb{N} $ - forms the building blocks of polynomials) $ p(x) $ is of degree $ n $ and the coefficient of the highest degree term $ {a_n} $ is positive.
Complete step by step answer:
To solve the given question, follow the facts about polynomials:
If $ p(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ... + {a_1}x + {a_0} $ is a polynomial of degree $ n $ then,
$ \mathop {\lim }\limits_{x \to \infty } p(x) = \mathop {\lim }\limits_{x \to \infty } {a_n}{x^n} $ and
$ {\lim _{x \to - \infty }}p(x) = {\lim _{x \to - \infty }}{a_n}{x^n} $
According to this fact, when we take a limit at infinity for a polynomial, all we have to do is look at the term with the largest power and ask what that term is doing in the limit.
If $ a $ is positive and $ n $ is even, then $ {a_n}{x^n} $ is always positive, and
$ {\lim _{x \to - \infty }}p(x) = {\lim _{x \to - \infty }}{a_n}{x^n} = \infty $
If $ a $ is positive and $ n $ is odd, then $ {a_n}{x^n} $ is negative when $ x $ is negative.
So
$\Rightarrow$ $ {\lim _{x \to - \infty }}p(x) = {\lim _{x \to - \infty }}{a_n}{x^n} = - \infty $
Note:
If the values of $ f(x) $ become arbitrarily close to L as $ x $ becomes sufficiently large, we say the function $ f $ has a limit at infinity and write
$ \mathop {\lim }\limits_{x \to \infty } f(x) = L $
If the values of $ f(x) $ becomes arbitrarily close to $ L $ for $ x \ll 0 $ as $ \left| x \right| $ becomes sufficiently large, we say that the function $ f $ has a limit at negative infinity and write
$ {\lim _{x \to - \infty }}f(x) = L $
If the values $ f(x) $ are getting arbitrarily close to some finite value $ L $ as $ x \to \infty $ or $ x \to - \infty $ , the graph of $ f $ approaches the line $ y = L $ .
In that case, the line $ y = L $ is a horizontal asymptote( a straight line that continually approaches a given curve but does not meet it at any finite distance) of $ f $ (figure).
Complete step by step answer:
To solve the given question, follow the facts about polynomials:
If $ p(x) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ... + {a_1}x + {a_0} $ is a polynomial of degree $ n $ then,
$ \mathop {\lim }\limits_{x \to \infty } p(x) = \mathop {\lim }\limits_{x \to \infty } {a_n}{x^n} $ and
$ {\lim _{x \to - \infty }}p(x) = {\lim _{x \to - \infty }}{a_n}{x^n} $
According to this fact, when we take a limit at infinity for a polynomial, all we have to do is look at the term with the largest power and ask what that term is doing in the limit.
If $ a $ is positive and $ n $ is even, then $ {a_n}{x^n} $ is always positive, and
$ {\lim _{x \to - \infty }}p(x) = {\lim _{x \to - \infty }}{a_n}{x^n} = \infty $
If $ a $ is positive and $ n $ is odd, then $ {a_n}{x^n} $ is negative when $ x $ is negative.
So
$\Rightarrow$ $ {\lim _{x \to - \infty }}p(x) = {\lim _{x \to - \infty }}{a_n}{x^n} = - \infty $
Note:
If the values of $ f(x) $ become arbitrarily close to L as $ x $ becomes sufficiently large, we say the function $ f $ has a limit at infinity and write
$ \mathop {\lim }\limits_{x \to \infty } f(x) = L $
If the values of $ f(x) $ becomes arbitrarily close to $ L $ for $ x \ll 0 $ as $ \left| x \right| $ becomes sufficiently large, we say that the function $ f $ has a limit at negative infinity and write
$ {\lim _{x \to - \infty }}f(x) = L $
If the values $ f(x) $ are getting arbitrarily close to some finite value $ L $ as $ x \to \infty $ or $ x \to - \infty $ , the graph of $ f $ approaches the line $ y = L $ .
In that case, the line $ y = L $ is a horizontal asymptote( a straight line that continually approaches a given curve but does not meet it at any finite distance) of $ f $ (figure).
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