Prove by limit definition

$$\lim _{x\to \infty }\left(\frac{-7x^2+9x}{4x^2+8}\right)=\frac{-7}{4}$$

let $\epsilon > 0$ need to find $M$ such that for every $x>M \implies |f(x) - L|<\epsilon$

$\left|\frac{-7x^2+9x}{4x^2+8}+\frac{7}{4}\right| = \frac{9x+14}{4\left(x^2+2\right)} \le \frac{23x}{4x^2}=\frac{23}{4x} < \epsilon $

so I choose $M=\frac{23}{4\epsilon}$

My question is that I assumed that $x>0$ do I need to check when $x\le0$ or this is enough since $x \to \infty $ ? and does this prove the limit ?


  • $\begingroup$ @KaviRamaMurthy $\left|\frac{-7x^2+9x}{4x^2+8}+\frac{7}{4}\right|=\left|\frac{4\left(-7x^2+9x\right)+7\left(4x^2+8\right)}{4\left(4x^2+8\right)}\right|=\left|\frac{-28x^2+36x+28x^2+56}{4\left(4x^2+8\right)}\right|=\left|\frac{36x+56}{4\left(4x^2+8\right)}\right|=\left|\frac{9x+14}{\left(4x^2+8\right)}\right|$ $\endgroup$ – user739993 Jan 7 at 9:14
  • $\begingroup$ Sorry, I was wrong. $\endgroup$ – Kavi Rama Murthy Jan 7 at 9:21
  • $\begingroup$ if is written the limit at $x\to\infty$, so it means you have to count limits at $x\to+\infty$ and $x\to-\infty$. $\endgroup$ – thing Jan 7 at 10:48
  • $\begingroup$ @thing You are wrong… $x\to\infty$ means $x \to +\infty$ $\endgroup$ – Gono Jan 7 at 11:32
  • $\begingroup$ No, you don't need to check for $x \le 0$. What you have done proves the limit $\endgroup$ – Norse Jan 7 at 11:34

Look for example at pp. 105, Calculus (Third Edition) from Spivak. The definition of $\lim_{x\to\infty} f(x)=L$ is that for every $\varepsilon>0$ there is a number $N$ such that, for all $x$, if $x>N$, then $|f(x)-L|<\varepsilon$. According to this, what you have done proves the limit, but I think that in your attempt to solve the problem it was necessary to assume $x>1$. Nevertheless this is not a problem because you can take $M=\max\{1, \frac{23}{4\varepsilon}\}$ to ensure that the bounds you took are ok.

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