Eve was tasked with calculating the following limit: $$ \lim_{x \rightarrow \infty}\frac{x^2+1}{2x^2} $$ From her math lessons, she remembered that if she is calculating the limit of a quotient and both the numerator and the denominator tend to infinity, she can use l’Hospital’s rule. Thanks to l’Hospital’s rule, she can use differentiation to find the limit.
(1) Since, in this case, it is a quotient, she applied the following differentiation rule: $$ \left(\frac{f}{g}\right)'=\frac{f' g-fg'}{g^2 } $$ and modified the limit as follows: $$ \lim_{x \rightarrow \infty} \frac{x^2+1}{2x^2 } =\lim_{x \rightarrow \infty} \frac{2x \cdot 2x ^2-(x^2+1) \cdot 4x}{4x^4 } $$
(2) Then, she simplified the expression in the limit: $$ \lim_{x \rightarrow \infty} \frac{ 2x \cdot 2x^2-(x^2+1)\cdot 4x}{4x^4} =\lim_{x \rightarrow \infty} \frac{4x ^3-4x^3-4x}{4x^4}=\lim_{x \rightarrow \infty} \frac{-4x}{4x^4}=\lim_{x \rightarrow \infty} \frac{-1}{x^3} $$ The problematic $x$, causing the expression to be of the form $\frac{\infty}{\infty}$, was cancelled and eliminated from the numerator.
(3) Further, in calculating this limit, Eve applied the fact that $$ \lim_{x \rightarrow \infty} \frac{1}{x} =0 $$ and got: $$ \lim_{x \rightarrow \infty} \frac{-1}{x^3} =0 $$
(4) Based on this result, she concluded that $$ \lim_{x \rightarrow \infty} \frac{x^2+1}{2x^2} =0. $$ Did Eve make a mistake? If yes, specify where.
Yes. The mistake is in step (1). L’Hospital’s rule is not applied correctly.
Yes. The mistake is in step (2). We cannot write $\frac{-4x}{4x^4}=\frac{-1}{x^3}$. We don’t know if $x=0$.
Yes. The mistake is in step (3). It should have been: $$\lim_{x \rightarrow \infty} \frac{-1}{x^3} =-\infty$$
Yes. All steps are unnecessary. Expression of the form $\frac{\infty}{\infty}$ always equals $1$.
No. The whole procedure is correct.
When applying L’Hospital’s rule, we must consider the fact that the numerator and denominator are differentiated separately. Eve confused this rule with the derivative of a quotient, so her procedure for finding the limit is not correct. The correct way to apply l’Hospital’s rule is: $$ \lim_{x \rightarrow \infty} \frac{x^2+1}{2x^2 } =\lim_{x \rightarrow \infty} \frac{2x}{4x}=\lim_{x \rightarrow \infty} \frac24=\frac12. $$