Step-by-step explanation:
When x = 0,
sin3x / 6x = 0 / 0, which is an indeterminate form.
Hence we use L'Hospital's rule:
d/dx (sin3x) = 3cos3x
d/dx (6x) = 6
Now we have 3cos3x / 6 or 0.5cos3x.
When x = 0, 0.5cos3x = 0.5(1) = 0.5.
Hence the limit is 0.5.
Using limits, it is found that the value of the function is of 0.5.
The limit given is:
[tex]\lim_{x \rightarrow 0} \frac{\sin{(3x)}}{6x}[/tex]
The following property is used:
[tex]\lim_{x \rightarrow 0} \frac{\sin{(ax)}}{ax} = 1[/tex]
Hence, multiplying the numerator and denominator by 0.5:
[tex]\lim_{x \rightarrow 0} \frac{\sin{(3x)}}{6x} = 0.5\lim_{x \rightarrow 0} \frac{\sin{(3x)}}{3x} = 0.5(1) = 0.5[/tex]
The value of the function is of 0.5.
To learn more about limits, you can take a look at https://brainly.com/question/24482719
