 maths > calculus-limits

Limit of Polynomials

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Limit of Polynomials

Apply Algebra of Limits
For a function $f\left(x\right)={a}_{n}{x}^{n}$$f \left(x\right) = {a}_{n} {x}^{n}$$+{a}_{n-1}{x}^{n-1}$$+ {a}_{n - 1} {x}^{n - 1}$$+\cdots$$+ \cdots$$+{a}_{1}{x}^{1}$$+ {a}_{1} {x}^{1}$$+{a}_{0}$$+ {a}_{0}$

$\underset{x\to a}{lim}f\left(x\right)$${\lim}_{x \to a} f \left(x\right)$

$\quad \quad = {a}_{n} {\lim}_{x \to a} {x}^{n}$$+{a}_{n-1}\underset{x\to a}{lim}{x}^{n-1}$$+ {a}_{n - 1} {\lim}_{x \to a} {x}^{n - 1}$$+\cdots$$+ \cdots$$+{a}_{1}\underset{x\to a}{lim}{x}^{1}$$+ {a}_{1} {\lim}_{x \to a} {x}^{1}$$+{a}_{0}$$+ {a}_{0}$

limit of a polynomial

Using the algebra of limits, the limit of function $f\left(x\right)=3{x}^{3}+2{x}^{2}-1$$f \left(x\right) = 3 {x}^{3} + 2 {x}^{2} - 1$ at $x=-2$$x = - 2$ is found by applying limit to the three terms of the function.

limit of function $f\left(x\right)=3{x}^{3}+2{x}^{2}-1$$f \left(x\right) = \textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{3}} + \textcolor{c \mathmr{and} a l}{2 {x}^{2}} - 1$ at $x=-2$$x = - 2$

By Substitution :
$f\left(x\right){\mid }_{x=-2}$$f \left(x\right) {|}_{x = - 2}$
$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{3 {\left(- 2\right)}^{3}} + \textcolor{c \mathmr{and} a l}{2 {\left(- 2\right)}^{2}} - 1$
$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{3 \left(- 8\right)} + \textcolor{c \mathmr{and} a l}{2 \left(4\right)} - 1$
$\quad \quad = - 17$

left-hand-limit :
$\underset{x\to -2-}{lim}f\left(x\right)$${\lim}_{x \to - 2 -} f \left(x\right)$
$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{3 {\left(- 2 - \delta\right)}^{3}} + \textcolor{c \mathmr{and} a l}{2 {\left(- 2 - \delta\right)}^{2}} - 1$
substitute $\delta =0$$\delta = 0$
$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{3 \left(- 8\right)} + \textcolor{c \mathmr{and} a l}{2 \left(4\right)} - 1$
$\quad \quad = - 17$

right-hand-limit :
$\underset{x\to -2+}{lim}f\left(x\right)$${\lim}_{x \to - 2 +} f \left(x\right)$
$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{3 {\left(- 2 + \delta\right)}^{3}} + \textcolor{c \mathmr{and} a l}{2 {\left(- 2 + \delta\right)}^{2}} - 1$
substitute $\delta =0$$\delta = 0$
$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{3 \left(- 8\right)} + \textcolor{c \mathmr{and} a l}{2 \left(4\right)} - 1$
$\quad \quad = - 17$

Summary :
limits of function $f\left(x\right)=3{x}^{3}+2{x}^{2}-1$$f \left(x\right) = 3 {x}^{3} + 2 {x}^{2} - 1$ at $x=-2$$x = - 2$

$f\left(x\right){\mid }_{x=-2}=-17$$f \left(x\right) {|}_{x = - 2} = - 17$

$\underset{x\to -2-}{lim}f\left(x\right)=-17$${\lim}_{x \to - 2 -} f \left(x\right) = - 17$

$\underset{x\to -2+}{lim}f\left(x\right)=-17$${\lim}_{x \to - 2 +} f \left(x\right) = - 17$

All three values are equal. So the function is continuous.

summary

Limit of a polynomial: For a function $f\left(x\right)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\cdots +{a}_{1}{x}^{1}+{a}_{0}$$f \left(x\right) = {a}_{n} {x}^{n} + {a}_{n - 1} {x}^{n - 1} + \cdots + {a}_{1} {x}^{1} + {a}_{0}$
$\underset{x\to a}{lim}f\left(x\right)$${\lim}_{x \to a} f \left(x\right)$
$\quad \quad = {a}_{n} {\lim}_{x \to a} {x}^{n} + {a}_{n - 1} {\lim}_{x \to a} {x}^{n - 1} + \cdots$
$\quad \quad \quad \quad + {a}_{1} {\lim}_{x \to a} {x}^{1} + {a}_{0}$

Outline