Tentukan nilai $\delta$ terbesar sehingga konsisten dengan definisi limit pada limit fungsi berikut : 1.$\displaystyle \lim_{x\rightarrow 1} \frac{x^2-1}{x-1}=2$ Kotretan : $|f(x)-L|=|\frac{x^2-1}{x-1}-2|=|\frac{x^2-1}{x-1}-\frac{2(x-1)}{(x-1)}|=|\frac{x^2-1-2x+2}{x-1}|=|\frac{x^2-1-2x+2}{x-1}|$ $=|\frac{x^2-2x+1}{x-1}|=|\frac{(x-1)(x-1)}{x-1}|=|x-1|<\delta=\epsilon$ Jadi，pilih $\delta=\epsilon$. Bukti : Ambil sebarang $\epsilon >0$. Pilih $\delta=\epsilon$, sedemikian sehingga jika kita misalkan $|x-1|<\delta$, maka $|f(x)-L|=|\frac{x^2-1}{x-1}-2|=|\frac{x^2-1}{x-1}-\frac{2(x-1)}{(x-1)}|=|\frac{x^2-1-2x+2}{x-1}|=|\frac{x^2-1-2x+2}{x-1}|=|\frac{x^2-2x+1}{x-1}|=|\frac{(x-1)(x-1)}{x-1}|=|x-1|<\delta=\epsilon$ 2.$\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x-1}=\frac{1}{2}$ Kotretan : $|f(x)-L|=|\frac{\sqrt{x}-1}{x-1}-\frac{1}{2}|=|\frac{\sqrt{x}-1}{x-1}-\frac{1(x-1)}{2(x-1)}|=|\frac{2(\sqrt{x}-1)-(x-1)}{2(x-1)}|=|\frac{2\sqrt{x}-2-x+1}{2(x-1)}|$ Kita harus memunculkan $|x-1|$ di ...