Respuesta :
If we evaluate the function at infinity, we get:
[tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle L = \lim_{x \to \infty}{ \frac{\sqrt{4x^4 + x^2 + 1}}{x^2 + 1} } = \frac{\infty}{\infty} } \end{gathered}$}[/tex]
Therefore, being an indeterminate form, we must multiply and divide by the highest degree monomial (x²; remember that we divide the powers of the numerator by 2 since they are inside a square root):
[tex]\begin{align*}L &= \lim_{x \to \infty}{ \frac{\sqrt{4x^4 + x^2 + 1}}{x^2 + 1} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}}\\ & = \lim_{x \to \infty}{\frac{ \sqrt{4 + \frac{1}{x^2} + \frac{1}{x^4}} }{1 + \frac{1}{x^2}}} \end{align*}[/tex][tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{L= \lim_{x \to \infty}\frac{\sqrt{4x^2+x^{2} +1 } }{x^2+1}\cdot\frac{\frac{1}{x^{2} } }{\frac{1}{x^{2} } } } \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf \bf{ \ \ \ = \lim_{x \to \infty} \frac{\sqrt{4+\frac{1}{x^{2} }+\frac{1}{x^{4} } } }{1+\frac{1}{x^{2} } } } \end{gathered}$}[/tex]
Then, evaluated at infinity, we have:
[tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle L = \frac{\sqrt{4 + 0 + 0}}{1 + 0} = \frac{\sqrt{4}}{1} = 2} \end{gathered}$}[/tex]
So the limit is 2.
