Can $\lim_{x \to \infty} = \frac{x^2}{x-1}$ be solved without L'hopital

Question

I need to find limit for: $$\lim_{x \to \infty} = \frac{x^2}{x-1}$$

I could solve this easily with L'hopital's rule. Taking derivative of both numerator and denumatorator gives:

$$\lim_{x \to \infty} = \frac{2x}{1} = \infty$$

But is there a way to solve this without using it? I tried multiplying both parts of the fraction with conjugate $(x+1)$, and then dividing with largest element. But I get zero in the denumerator.

Answer 1

Yes, there is, and for such simple problems, L’Hopital’s rule should never be used. When dealing with limits of rational functions (polynomial divided by polynomial), you always factor the highest powers from the numerator and denominator: \begin{align} \frac{x^2}{x-1}=\frac{x^2\cdot(1)}{x\cdot\left(1-\frac{1}{x}\right)}=x\cdot\frac{1}{1-\frac{1}{x}}. \end{align} As $x\to\infty$, the first term which is just $x$ obviously approaches $\infty$. The fraction on the other hand approaches $\frac{1}{1-0}=1$. Hence, their product approaches $\infty$.

This reasoning will show you more generally that if $f(x)=\frac{a_nx^n+\dots +a_0}{b_mx^m+\cdots +b_0}$ is a rational function with $a_n,b_m\neq 0$, then

• $n>m$ implies that $ \lim\limits_{x\to\infty}f(x)=\begin{cases} \infty &\text{if $\frac{a_n}{b_m}>0$}\\ -\infty & \text{if $\frac{a_n}{b_m}<0$}. \end{cases} $
• $n=m$ implies that $\lim\limits_{x\to\infty}f(x)=\frac{a_n}{b_n}$.
• $n<m$ implies that $\lim\limits_{x\to\infty}f(x)=0$.

Answer 2

It can be graphically understood, $f(x)=x^2$ is a highly increasing function for larger values of $x$ [since $f'(x)=2x$, $x$ increases $f'(x)$ increases]

Whereas $g(x)=x-1$ has constant $g'(x)=1$

Therefore when $x$ tends to infinity the expression $f(x)/g(x)$ tends to infinity

Answer 3

If we multiply the numerator and denominator by the reciprocal of the highest power of $x$ in the denominator, we obtain \begin{align*} \lim_{x \to \infty} \frac{x^2}{x - 1} & = \lim_{x \to \infty} \frac{x^2}{x - 1} \cdot \frac{\frac{1}{x}}{\frac{1}{x}}\\ & = \lim_{x \to \infty} \frac{x}{1 - \frac{1}{x}}\\ & = \infty \end{align*} since the numerator grows without bound while the denominator approaches $1$.

Alternatively, observe that $$\frac{x^2}{x - 1} = x + 1 + \frac{1}{x - 1} > x + 1$$ whenever $x > 1$. Hence, $$\lim_{x \to \infty} \frac{x^2}{x - 1} \geq \lim_{x \to \infty} (x + 1) = \infty$$

Answer 4

Note that: $$\frac{x^{2}}{x-1} = \frac{x^{2}-1+1}{x-1} = \frac{x^{2}-1}{x-1}+\frac{1}{x-1} = \frac{(x-1)(x+1)}{x-1} + \frac{1}{x-1} = x+1 + \frac{1}{x-1}.$$ The first limit goes to infinity and the second goes to zero, so the result is: $$\lim_{x\to +\infty}\frac{x^{2}}{x-1} = +\infty.$$

Answer 5

Choose some value of N. Assume that N is large (like at least 7).

If $x>N$ then $\frac {x^2}{x-1} > N$

Now assume that N is even bigger than that! The previous statement still holds. And, it holds for $N$ arbitrarily large.