I want to prove that $\lim\limits_{x\to -\infty} 10-\sqrt{4-x} = -\infty$ using this definition: $$(\forall M\in\mathbb{R})(\exists N\in\mathbb{R})(\forall x\in\mathbb{R})[(x<N)\rightarrow (10-\sqrt{4-x}<M)]$$ But I am having trouble coming up with a suitable $N$ value. This is what I have tried and why I have failed in trying to make $f(x)<M$:
I have tried letting $N=-\Big(10-\frac{1}{M}\Big)^2$, but this does not work since $\frac{1}{M} > M$ for $M < 0$ and $M$ could not equal zero.
I have tried the inequality $10-\sqrt{4-x}<10-\sqrt{-x}<10-\frac{1}{x}<M$, but $M$ would have to be greater than 10 for $N=\frac{1}{10-M}$ to work.
Any ideas on how to solve this problem using only the definition?
Thanks!
Given $M$, let $N=4-(M-10)^2$. With this choice, if we have $x<N$, then $$4-x>4-N=(M-10)^2,$$ hence $$\sqrt{4-x}> |M-10|=\max\{M-10,10-M\}\ge 10-M$$ and $$10-\sqrt{4-x}< 10-(10-M)=M.$$