iv. evaluate the limit of the function\na. $lim_{x\rightarrow2}3x^{3}-2x^{2}+x - 4$\nb. $lim_{x\rightarrow…

Explanation:

Step1: Recall limit - laws for polynomials

For a polynomial function (y = f(x)=a_nx^n+\cdots + a_1x + a_0), (\lim_{x\rightarrow c}f(x)=f(c)) when (f(x)) is continuous at (x = c).

Step2: Evaluate (\lim_{x\rightarrow 2}(3x^{3}-2x^{2}+x - 4))

Substitute (x = 2) into the polynomial: [ \begin{align*} &3(2)^{3}-2(2)^{2}+2 - 4\ =&3\times8-2\times4 + 2-4\ =&24-8 + 2-4\ =&14 \end{align*} ]

Step3: Evaluate (\lim_{x\rightarrow - 1}\sqrt{x^{2}+3x - 1})

First, check the value of the expression inside the square - root when (x=-1): ((-1)^{2}+3(-1)-1=1 - 3-1=-3). Since the expression inside the square - root is negative, the limit is not a real - valued number in the set of real numbers.

Step4: Evaluate (\lim_{x\rightarrow1}\frac{2x^{3}-x^{2}-x}{x^{2}+3x - 2}-4)

First, find (\lim_{x\rightarrow1}\frac{2x^{3}-x^{2}-x}{x^{2}+3x - 2}). Substitute (x = 1) into the rational function: (\frac{2(1)^{3}-(1)^{2}-1}{(1)^{2}+3(1)-2}=\frac{2 - 1-1}{1 + 3-2}=0). Then (\lim_{x\rightarrow1}\frac{2x^{3}-x^{2}-x}{x^{2}+3x - 2}-4=0 - 4=-4).

Step5: Evaluate (\lim_{x\rightarrow2}\sqrt{3x - 2})

Substitute (x = 2) into the function: (\sqrt{3(2)-2}=\sqrt{4}=2).

Step6: Evaluate (\lim_{x\rightarrow0^{+}}\frac{3}{1 + 2^{\frac{1}{x}}})

As (x\rightarrow0^{+}), (\frac{1}{x}\rightarrow+\infty), and (2^{\frac{1}{x}}\rightarrow+\infty). Then (1 + 2^{\frac{1}{x}}\rightarrow+\infty), so (\lim_{x\rightarrow0^{+}}\frac{3}{1 + 2^{\frac{1}{x}}}=0).

Step7: Evaluate (\lim_{x\rightarrow0^{-}}\frac{3}{1 + 2^{\frac{1}{x}}})

As (x\rightarrow0^{-}), (\frac{1}{x}\rightarrow-\infty), and (2^{\frac{1}{x}}\rightarrow0). Then (\lim_{x\rightarrow0^{-}}\frac{3}{1 + 2^{\frac{1}{x}}}=\frac{3}{1+0}=3).

Step8: Evaluate (\lim_{x\rightarrow+\infty}\frac{3x^{3}+2x - 3}{2x^{3}-3})

Divide both the numerator and denominator by (x^{3}): (\lim_{x\rightarrow+\infty}\frac{3+\frac{2}{x^{2}}-\frac{3}{x^{3}}}{2-\frac{3}{x^{3}}}). As (x\rightarrow+\infty), (\frac{2}{x^{2}}\rightarrow0) and (\frac{3}{x^{3}}\rightarrow0). So (\lim_{x\rightarrow+\infty}\frac{3x^{3}+2x - 3}{2x^{3}-3}=\frac{3}{2}).

Step9: Evaluate (\lim_{x\rightarrow+\infty}(x+\frac{3}{x^{2}}+\frac{3}{x^{3}}))

As (x\rightarrow+\infty), (\frac{3}{x^{2}}\rightarrow0) and (\frac{3}{x^{3}}\rightarrow0), so (\lim_{x\rightarrow+\infty}(x+\frac{3}{x^{2}}+\frac{3}{x^{3}})=+\infty).

Step10: Evaluate (\lim_{x\rightarrow0}\frac{\tan x}{x})

We know that (\lim_{x\rightarrow0}\frac{\tan x}{x}=\lim_{x\rightarrow0}\frac{\sin x}{x}\cdot\frac{1}{\cos x}). Since (\lim_{x\rightarrow0}\frac{\sin x}{x}=1) and (\lim_{x\rightarrow0}\cos x = 1), then (\lim_{x\rightarrow0}\frac{\tan x}{x}=1).

Answer:

a. (14) b. Not a real - valued number c. (-4) d. (2) e. (0) (for (x\rightarrow0^{+})), (3) (for (x\rightarrow0^{-})) f. Does not exist (left - hand limit and right - hand limit are different) g. (\frac{3}{2}) h. (+\infty) i. (1)