Prove $\lim_{x\to 7/4^+}\tfrac{3x}{4x-7}=\infty$ by definition $$\lim_{x\to \frac{7}{4}^+}\frac{3x}{4x-7}=\infty $$

I want to prove that for every $M>0$ exists $\delta$ for which $ 0<x-\frac{7}{4}<\delta $  such that $ f(x)>M $  

What I tried:
$$f(x)=\frac{3x}{4x-7}>M\iff\frac{1}{M}>\frac{4x-7}{3x} =* $$
Then I took some $\delta_1 = \frac{1}{4}$ and then:
$$0<x-\frac{7}{4}<\frac{1}{4}\iff \frac{7}{4}<x<2\iff7<4x<8\iff0<4x-7<1 $$ 

Final proof:
We take arbitrary $M>0 $.
Then we take some  $ \delta_{1}=\frac{1}{4}$. Then
$$ 0<x-\cfrac{7}{4}<\cfrac{1}{4}\iff \cfrac{7}{4}<x<2\leftrightarrow\cfrac{21}{4}<3x<6 $$
So $$ f(x)=\cfrac{3x}{4x-7}>\cfrac{5}{4x-7}=\cfrac{5}{4\left(x-\cfrac{7}{4}\right)}=\cfrac{5}{4}\cdot\cfrac{1}{x-\cfrac{7}{4}}>M\iff \cfrac{1}{x-\cfrac{7}{4}}>\cfrac{4M}{5}\iff x-\cfrac{7}{4}<\frac{5}{4M}$$
Then $ \delta_{2}=\frac{5}{4M} $
$$ \delta=\min\{\delta_{1},\delta_{2}\}=\min\left\{\cfrac{5}{4M},\cfrac{1}{8}\right\} $$

The part that makes the limit go to infinity is the denominator, the numerator will be larger than 3*7/4 > 5.
For x>7/4 we have $$f(x) > \frac{5}{4x-7} = \frac{5}{4} \cdot \frac{1}{x-7/4} > M$$ 
This holds if
$$ 0 < x-7/4 < \frac{5}{4M} = \delta$$
