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Let $f(x) = \sqrt{x^2-5x+1}-x$

Find $\lim_{x\to\infty}f(x)$

$$\lim_{x\to\infty} \sqrt{x^2-5x+1}-x$$

$$\lim_{x\to\infty} \dfrac{x^2-5x+1-x^2}{\sqrt{x^2-5x+1}+x}$$

$$\lim_{x\to\infty} \dfrac{-5x+1}{\sqrt{x^2-5x+1}+x}$$

$$\lim_{x\to\infty} \dfrac{\dfrac{-5x+1}{x}}{\dfrac{\sqrt{x^2-5x+1}+x}{x}}$$

$$\lim_{x\to\infty} \dfrac{-5+\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{5}{x}+\dfrac{1}{x^2}}+1}$$

From here, I know that $\lim_{x\to\infty} \dfrac{1}{x} = 0$, $\lim_{x\to\infty} \dfrac{5}{x} = 0$ and $\lim_{x\to\infty} \dfrac{1}{x^2} = 0$

$$\lim_{x\to\infty} \dfrac{-5}{\sqrt{1}+1}$$

$$\lim_{x\to\infty} \dfrac{-5}{2}$$

$$\lim_{x\to\infty} f(x) = \dfrac{-5}{2}$$

Everything up to here seems fine. The issue is when I try to find $\lim_{x\to-\infty} f(x)$

I also know that $\lim_{x\to-\infty} \dfrac{1}{x} = 0$, $\lim_{x\to-\infty} \dfrac{5}{x} = 0$ and $\lim_{x\to-\infty} \dfrac{1}{x^2} = 0$

This would make me conclude that $\lim_{x\to-\infty}f(x) = \dfrac{-5}{2}$.

However, this is not the case because $\lim_{x\to-\infty}f(x) = \infty$

Desmos view of $f(x)$

Why am I arriving to the wrong answer and how can I algebraically prove that the answer is $\infty$?

Cedric Martens
  • 1,288

3 Answers3

4

hint

At $-\infty$, $x$ becomes negative, thus

$$\sqrt{x^2}=|x|=-x$$

$$\frac{\sqrt{x^2-5x+1}}{x}=\frac{\sqrt{x^2-5x+1}}{-(-x)}=$$

$$-\sqrt{\frac{x^2-5x+1}{x^2}}.$$

the denominator goes to zero.

hamam_Abdallah
  • 62,951
3

Note that for $x\to -\infty$

  • $\sqrt{x^2-5x+1} \to +\infty$
  • $-x \to +\infty$

and therefore

$$\sqrt{x^2-5x+1}-x \to +\infty$$

As an alternative by $x=-y\to -\infty$ with $y\to +\infty$ we have

$$\lim_{x\to-\infty}\sqrt{x^2-5x+1}-x=\lim_{y\to +\infty}\sqrt{y^2+5y+1}+y \to +\infty$$

user
  • 154,566
2

as far as understanding, you have, in essence, $$ \left| x - \frac{5}{2} \right| - x . $$ for large positive $x$ you get close to $-5/2,$ but for large negative $x,$ meaning $x$ is negative and $|x|$ is large, you have roughly $$ 2 |x| + \frac{5}{2} $$ which grows without bound. For example, if $x = -10,$ the original expression becomes $$ \sqrt{100 - (-50) + 1} - (-10) = \sqrt {151} + 10 \approx 22.2882 $$ If $x = -100,$ $$ \sqrt{10000 - (-500) + 1} - (-100) = \sqrt {10501} + 100 \approx 202.474 $$ If $x = -1000,$ $$ \sqrt{1000000 - (-5000) + 1} - (-1000) = \sqrt {1005001} + 1000 \approx 2002.49738 $$

Will Jagy
  • 139,541
  • Thank you! Is there a way to quickly see that the result is unbounded without jumping to conclusions on the wrong answer like I did? – Cedric Martens Sep 22 '18 at 21:13
  • @CedricMartens probably not quick. Just that the square root term amounts, more or less, to an absolute value. Think about $|x| - x,$ how could you express that without any absolute value signs in case (I) $x > 0$ ? but then case (II) $x < 0$ ?? – Will Jagy Sep 22 '18 at 21:20