Question

iv. evaluate the limit of the function
a. $lim_{x
ightarrow2}3x^{3}-2x^{2}+x - 4$
b. $lim_{x
ightarrow - 1}sqrt{x^{2}+3x - 1}$
c. $lim_{x
ightarrow1}\frac{2x^{3}-x^{2}-x}{x^{2}+3x - 2}-4$
d. $lim_{x
ightarrow2}sqrt{3x - 2}$
e. $lim_{x
ightarrow0^{+}}\frac{3}{1 + 2^{\frac{1}{x}}}$
f. $lim_{x
ightarrow0^{-}}\frac{3}{1+2^{\frac{1}{x}}}$
g. $lim_{x
ightarrowinfty}\frac{3x^{2}+2x - 3}{2x^{2}-3}$
h. $lim_{x
ightarrowinfty}x+\frac{3}{x^{2}}-\frac{3}{x^{3}}$
i. $lim_{x
ightarrow0}\frac{\tan x}{x}$

Explanation:

Step1: Recall limit - laws for polynomials

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

Step2: Evaluate \(\lim_{x

ightarrow 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

ightarrow - 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

ightarrow1}\frac{2x^{3}-x^{2}-x}{x^{2}+3x - 2}-4\)
First, find \(\lim_{x
ightarrow1}\frac{2x^{3}-x^{2}-x}{x^{2}+3x - 2}\). 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 \(\lim_{x
ightarrow1}\frac{2x^{3}-x^{2}-x}{x^{2}+3x - 2}-4=0 - 4=-4\).

Step5: Evaluate \(\lim_{x

ightarrow2}\sqrt{3x - 2}\)
Substitute \(x = 2\) into the function: \(\sqrt{3(2)-2}=\sqrt{4}=2\).

Step6: Evaluate \(\lim_{x

ightarrow0^{+}}\frac{3}{1 + 2^{\frac{1}{x}}}\)
As \(x
ightarrow0^{+}\), \(\frac{1}{x}
ightarrow+\infty\), and \(2^{\frac{1}{x}}
ightarrow+\infty\). Then \(1 + 2^{\frac{1}{x}}
ightarrow+\infty\), so \(\lim_{x
ightarrow0^{+}}\frac{3}{1 + 2^{\frac{1}{x}}}=0\).

Step7: Evaluate \(\lim_{x

ightarrow0^{-}}\frac{3}{1 + 2^{\frac{1}{x}}}\)
As \(x
ightarrow0^{-}\), \(\frac{1}{x}
ightarrow-\infty\), and \(2^{\frac{1}{x}}
ightarrow0\). Then \(\lim_{x
ightarrow0^{-}}\frac{3}{1 + 2^{\frac{1}{x}}}=\frac{3}{1+0}=3\).

Step8: Evaluate \(\lim_{x

ightarrow+\infty}\frac{3x^{3}+2x - 3}{2x^{3}-3}\)
Divide both the numerator and denominator by \(x^{3}\): \(\lim_{x
ightarrow+\infty}\frac{3+\frac{2}{x^{2}}-\frac{3}{x^{3}}}{2-\frac{3}{x^{3}}}\). As \(x
ightarrow+\infty\), \(\frac{2}{x^{2}}
ightarrow0\) and \(\frac{3}{x^{3}}
ightarrow0\). So \(\lim_{x
ightarrow+\infty}\frac{3x^{3}+2x - 3}{2x^{3}-3}=\frac{3}{2}\).

Step9: Evaluate \(\lim_{x

ightarrow+\infty}(x+\frac{3}{x^{2}}+\frac{3}{x^{3}})\)
As \(x
ightarrow+\infty\), \(\frac{3}{x^{2}}
ightarrow0\) and \(\frac{3}{x^{3}}
ightarrow0\), so \(\lim_{x
ightarrow+\infty}(x+\frac{3}{x^{2}}+\frac{3}{x^{3}})=+\infty\).

Step10: Evaluate \(\lim_{x

ightarrow0}\frac{\tan x}{x}\)
We know that \(\lim_{x
ightarrow0}\frac{\tan x}{x}=\lim_{x
ightarrow0}\frac{\sin x}{x}\cdot\frac{1}{\cos x}\). Since \(\lim_{x
ightarrow0}\frac{\sin x}{x}=1\) and \(\lim_{x
ightarrow0}\cos x = 1\), then \(\lim_{x
ightarrow0}\frac{\tan x}{x}=1\).

Answer:

a. \(14\)
b. Not a real - valued number
c. \(-4\)
d. \(2\)
e. \(0\) (for \(x
ightarrow0^{+}\)), \(3\) (for \(x
ightarrow0^{-}\))
f. Does not exist (left - hand limit and right - hand limit are different)
g. \(\frac{3}{2}\)
h. \(+\infty\)
i. \(1\)