question 4\nevaluate the limit of a continuous fourth - degree polynomial at x = 1x = 1:\na)…

question 4\nevaluate the limit of a continuous fourth - degree polynomial at x = 1x = 1:\na) $lim_{x\rightarrow1}(x^{4}-5x^{3}+7x^{2}-3x + 1)$\nb) $lim_{x\rightarrow2}(x^{2}-5x^{3})$\nc) $lim_{x\rightarrow2}\frac{x^{3}+x^{2}-6x}{x^{2}-4}$\nd) $lim_{x\rightarrow3}\frac{x^{2}-9}{x - 3}$\ne) $lim_{x\rightarrow3}\frac{x^{2}-9}{x - 2}$\nf) $lim_{x\rightarrow2}f(x)$ where $f(x)=\begin{cases}x^{3}-4&\text{if }x < 2\\6&\text{if }x = 2\\2x&\text{if }x>2end{cases}$\ng) let $f(x)=\begin{cases}10 - x - x^{2}&\text{if }xleq4\\2x - 3&\text{if }x>4end{cases}$ calculate the following limits. enter \dne\ if the limit does not exist. $lim_{x\rightarrow4^{-}}f(x)=$ $lim_{x\rightarrow4^{+}}f(x)=$ $lim_{x\rightarrow4}f(x)=$

question 4\nevaluate the limit of a continuous fourth - degree polynomial at x = 1x = 1:\na) $lim_{x\rightarrow1}(x^{4}-5x^{3}+7x^{2}-3x + 1)$\nb) $lim_{x\rightarrow2}(x^{2}-5x^{3})$\nc) $lim_{x\rightarrow2}\frac{x^{3}+x^{2}-6x}{x^{2}-4}$\nd) $lim_{x\rightarrow3}\frac{x^{2}-9}{x - 3}$\ne) $lim_{x\rightarrow3}\frac{x^{2}-9}{x - 2}$\nf) $lim_{x\rightarrow2}f(x)$ where $f(x)=\begin{cases}x^{3}-4&\text{if }x < 2\\6&\text{if }x = 2\\2x&\text{if }x>2end{cases}$\ng) let $f(x)=\begin{cases}10 - x - x^{2}&\text{if }xleq4\\2x - 3&\text{if }x>4end{cases}$ calculate the following limits. enter \dne\ if the limit does not exist. $lim_{x\rightarrow4^{-}}f(x)=$ $lim_{x\rightarrow4^{+}}f(x)=$ $lim_{x\rightarrow4}f(x)=$

Answer

Explanation:

Step1: Evaluate limit a

For $\lim_{x\rightarrow1}(x^{4}-5x^{3}+7x^{2}-3x + 1)$, substitute $x = 1$ into the polynomial. $1^{4}-5\times1^{3}+7\times1^{2}-3\times1 + 1=1 - 5+7 - 3+1=1$

Step2: Evaluate limit b

For $\lim_{x\rightarrow2}(x^{2}-5x^{3})$, substitute $x = 2$ into the polynomial. $2^{2}-5\times2^{3}=4-5\times8=4 - 40=-36$

Step3: Evaluate limit c

For $\lim_{x\rightarrow2}\frac{x^{3}+x^{2}-6x}{x^{2}-4}$, factor the numerator and denominator. $x^{3}+x^{2}-6x=x(x^{2}+x - 6)=x(x + 3)(x - 2)$ and $x^{2}-4=(x + 2)(x - 2)$. Then $\lim_{x\rightarrow2}\frac{x(x + 3)(x - 2)}{(x + 2)(x - 2)}=\lim_{x\rightarrow2}\frac{x(x + 3)}{x + 2}=\frac{2\times(2 + 3)}{2+2}=\frac{10}{4}=\frac{5}{2}$

Step4: Evaluate limit d

For $\lim_{x\rightarrow3}\frac{x^{2}-9}{x - 3}$, factor the numerator. $x^{2}-9=(x + 3)(x - 3)$. Then $\lim_{x\rightarrow3}\frac{(x + 3)(x - 3)}{x - 3}=\lim_{x\rightarrow3}(x + 3)=6$

Step5: Evaluate limit e

For $\lim_{x\rightarrow3}\frac{x^{2}-9}{x - 2}$, factor the numerator. $x^{2}-9=(x + 3)(x - 3)$. Then substitute $x = 3$: $\frac{3^{2}-9}{3 - 2}=\frac{9 - 9}{1}=0$

Step6: Evaluate limit f

Find the left - hand limit: $\lim_{x\rightarrow2^{-}}f(x)=\lim_{x\rightarrow2^{-}}(x^{3}-4)=2^{3}-4=8 - 4 = 4$. Find the right - hand limit: $\lim_{x\rightarrow2^{+}}f(x)=\lim_{x\rightarrow2^{+}}(2x)=2\times2 = 4$. Since $\lim_{x\rightarrow2^{-}}f(x)=\lim_{x\rightarrow2^{+}}f(x)=4$, $\lim_{x\rightarrow2}f(x)=4$

Step7: Evaluate limit g (left - hand limit)

For $\lim_{x\rightarrow4^{-}}f(x)$, since $x\rightarrow4^{-}$ (approaching 4 from the left), use $f(x)=10 - x - x^{2}$. Substitute $x = 4$: $10-4 - 4^{2}=10-4 - 16=-10$

Step8: Evaluate limit g (right - hand limit)

For $\lim_{x\rightarrow4^{+}}f(x)$, since $x\rightarrow4^{+}$ (approaching 4 from the right), use $f(x)=2x - 3$. Substitute $x = 4$: $2\times4-3=5$

Step9: Evaluate limit g (overall limit)

Since $\lim_{x\rightarrow4^{-}}f(x)=-10$ and $\lim_{x\rightarrow4^{+}}f(x)=5$, $\lim_{x\rightarrow4}f(x)$ DNE

Answer:

a) $1$ b) $-36$ c) $\frac{5}{2}$ d) $6$ e) $0$ f) $4$ g) $\lim_{x\rightarrow4^{-}}f(x)=-10$, $\lim_{x\rightarrow4^{+}}f(x)=5$, $\lim_{x\rightarrow4}f(x)$ DNE