Question

Evaluate the following limit:


`\bf{\displaystyle \lim_(x \to \infty){ (√(4x^4 + x^2 + 1))/(x^2 + 1) }}`

Answer 1

If we evaluate the function at infinity, we get:

`\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle L = \lim_(x \to \infty){ (√(4x^4 + x^2 + 1))/(x^2 + 1) } = (\infty)/(\infty) } \end{gathered}$}`

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):

`\begin{align*}L &= \lim_(x \to \infty){ (√(4x^4 + x^2 + 1))/(x^2 + 1) \cdot ((1)/(x^2))/((1)/(x^2))}\\ & = \lim_(x \to \infty){\frac{ \sqrt{4 + (1)/(x^2) + (1)/(x^4)} }{1 + (1)/(x^2)}} \end{align*}`
`\large\displaystyle\text{$\begin{gathered}\sf \bf{L= \lim_(x \to \infty)\frac{\sqrt{4x^2+x^(2) +1 } }{x^2+1}\cdot((1)/(x^(2) ) )/((1)/(x^(2) ) ) } \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf \bf{ \ \ \ = \lim_(x \to \infty) \frac{\sqrt{4+(1)/(x^(2) )+(1)/(x^(4) ) } }{1+(1)/(x^(2) ) } } \end{gathered}$}`

Then, evaluated at infinity, we have:

`\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle L = (√(4 + 0 + 0))/(1 + 0) = (√(4))/(1) = 2} \end{gathered}$}`

So the limit is 2.