### Problem 8:
Determine whether the function [tex]\( h(x)=x^3-2x \)[/tex] is even, odd, or neither.
Solution:
Recall that a function [tex]\( f \)[/tex] is said to be even if [tex]\( f(-x) = f(x) \)[/tex] and odd if [tex]\( f(-x) = -f(x) \)[/tex].
Let's first evaluate [tex]\( h(-x) \)[/tex].
[tex]\[ h(-x) = (-x)^3 - 2(-x) \][/tex]
[tex]\[ h(-x) = -x^3 + 2x \][/tex]
Now, compare [tex]\( h(-x) \)[/tex] with [tex]\( h(x) \)[/tex]:
[tex]\[ h(x) = x^3 - 2x \][/tex]
Check for evenness:
[tex]\[ h(-x) \neq h(x) \][/tex]
As [tex]\( -x^3 + 2x \neq x^3 - 2x \)[/tex], the function is not even.
Check for oddness:
[tex]\[ -h(x) = - (x^3 - 2x) \][/tex]
[tex]\[ -h(x) = -x^3 + 2x \][/tex]
[tex]\[ h(-x) = -h(x) \][/tex]
Since [tex]\( h(-x) = -h(x) \)[/tex], the function is odd.
Hence, we see that [tex]\( h(-x) = -h(x) \)[/tex] and we conclude that [tex]\( h(x) \)[/tex] is odd.
### Problem 9:
Find the limit of the function [tex]\( f(x)=\frac{e^{3x}-1}{x} \)[/tex] as [tex]\( x \rightarrow 0 \)[/tex].
Solution:
Observe that as [tex]\( x \rightarrow 0 \)[/tex],
[tex]\[ \frac{e^{3x} - 1}{x} \][/tex]
becomes an indeterminate form [tex]\(\frac{0}{0}\)[/tex].
Therefore, we apply L'Hôpital's rule, which means differentiating the numerator and the denominator separately and then taking the limit of the new function.
First, differentiate the numerator and the denominator:
[tex]\[ \text{Numerator: } \frac{d}{dx} (e^{3x} - 1) = 3e^{3x} \][/tex]
[tex]\[ \text{Denominator: } \frac{d}{dx} x = 1 \][/tex]
Now, apply L'Hôpital's rule:
[tex]\[ \lim_{x \rightarrow 0} \frac{e^{3x} - 1}{x} = \lim_{x \rightarrow 0} \frac{3e^{3x}}{1} \][/tex]
Evaluate the limit:
[tex]\[ \lim_{x \rightarrow 0} 3e^{3x} = 3e^0 = 3 \][/tex]
Thus, the limit is:
[tex]\[ \boxed{3} \][/tex]
So, the limit of the function [tex]\( f(x) = \frac{e^{3x} - 1}{x} \)[/tex] as [tex]\( x \rightarrow 0 \)[/tex] is 3.
