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

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 ?

thanks

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