Answer:
The limit of the function exists
The value of the limit is 1/8
Explanations:
Given the limit of the function:
[tex]\lim _{x\to1}(\frac{1+3x}{1+4x^2+3x^4})^3[/tex]
In order to get the limit of the function, we will substitute x = 1 into the function;
[tex]\begin{gathered} =(\frac{1+3(1)}{1+4(1)^2+3(1)^4})^3 \\ =(\frac{1+3}{1+4(1)+3(1)})^3 \\ =(\frac{4}{1+4+3})^3 \\ =(\frac{4}{8})^3 \\ =(\frac{1}{2})^3 \\ =\frac{1}{8} \end{gathered}[/tex]
Since the limit of the function results in a finite value, hence the limit exists.
The value of the limit is 1/8
