Proving $ \lim_{x \to \infty } \frac{5x^{2}+1}{7x^{2}-3} = \frac{5}{7} $ using the limit definition

Question

I have problem particularly with this:

$$ \lim_{x \to \infty } \frac{5x^{2}+1}{7x^{2}-3} = \frac{5}{7} $$

I understand the logic of such proofs when it comes to simple functions as $f(x) = x + 4$ or $f(x) = \frac 1x$, however, I struggle with solving something like this.

Answer 1

Highlights:

$$\left|\frac{5x^2+1}{7x^2-3}-\frac57\right|=\frac{22}{7(7x^2-3)}\;\;(**)$$

You can take out the absolute value as $\;x\to\infty\;$ and thus you can assume $\;x>\sqrt\frac37\;$ . Also, for $\;x\;$ big enough as to fulfill $\;7x^2-3>22\;$ , say, you get above that

$$(**)<\frac17$$

Now get into the game $\;\epsilon,\,\delta\;$ and prove what you want.

Answer 2

I understand the logic of such proofs when it comes to simple functions as [...] f(x) = 1/x

Here is one way to reduce it to the case you know.

• First, write it as $\,f(x)=\frac{5x^{2}+1}{7x^{2}-3}=\frac{5}{7}\cdot\frac{x^2+1/5}{x^2-3/7}=\frac{5}{7}\cdot\left(1+\frac{22/35}{x^2-3/7}\right)=\frac{5}{7} \cdot \left(1+g(x)\right)\,$

• $g(x)$ only depends on $y=x^2$ so look at $g(x)=h(y)=\frac{22}{35}\cdot\frac{1}{y - 3/7}$

• Let $z=y-3/7$, then $h(y)=k(z)=\frac{22}{35}\cdot\frac{1}{z}$

Now, you know that $\lim_{z \to \infty} k(z) = 0\,$ from what you wrote. It follows from there that $\lim_{y \to \infty} h(y) = 0\,$, then $\lim_{x \to \infty} g(x) = 0\,$, and finally $\lim_{x \to \infty} f(x) = \frac{5}{7}\,$.