Question

How do you find the limit of \[\dfrac{{4{x^2} - 3x + 2}}{{7{x^2} + 2x - 1}}\] as \[x\] approaches infinity?

Answer 1

Hint: Here, the limit is defined as to prove the equation by evaluating the quotient rule, where the approaches to infinity. Let’s do the following limit rule to simplify the given equation by quotient rule of limit.

Complete step by step solution:
Given, \[\mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{4{x^2} - 3x + 2}}{{7{x^2} + 2x - 1}}\]
By dividing the numerator and denominator of \[f(x)\] with respect to highest exponent of \[x\] is \[{x^2}\] , we get
 \[
  f(x) = \dfrac{{4{x^2} - 3x + 2}}{{7{x^2} + 2x - 1}} \\
   = \dfrac{{\dfrac{{4{x^2}}}{{{x^2}}} - \dfrac{{3x}}{{{x^2}}} + \dfrac{2}{{{x^2}}}}}{{\dfrac{{7{x^2}}}{{{x^2}}} + \dfrac{{2x}}{{{x^2}}} - \dfrac{1}{{{x^2}}}}} \;
\]
By simplify, we get
 \[f(x) = \dfrac{{4 - \dfrac{3}{x} + \dfrac{2}{{{x^2}}}}}{{7 + \dfrac{2}{x} - \dfrac{1}{{{x^2}}}}}\]
Apply the limit of \[x\] approaches \[\infty \] becomes \[0\] , we get
 \[\mathop {\lim }\limits_{x \to \infty } f(x) = \dfrac{{4 - \dfrac{3}{\infty } + \dfrac{2}{{{\infty ^2}}}}}{{7 + \dfrac{2}{\infty } - \dfrac{1}{{{\infty ^2}}}}}\]
Where, \[\dfrac{1}{\infty } = 0\] so, we get
 \[\mathop {\lim }\limits_{x \to \infty } f(x) = \dfrac{{4 - 0 + 0}}{{7 + 0 - 0}}\]
Now, we have
 \[\mathop {\lim }\limits_{x \to \infty } f(x) = \dfrac{4}{7}\]
Hence, The value of \[\mathop {\lim }\limits_{x \to \infty } f(x) = \dfrac{4}{7}\] .
So, the correct answer is “$ \dfrac{4}{7}$”.

Note: We need to find the limit of the given function is in the form of quotient rule as approaches infinity by apply some methods to solve the problem and we remind the value \[\dfrac{1}{\infty } = 0\] to solve this type of limit problem.