a, $\lim_{x \to5} \frac{x-5}{\sqrt[]{x}-\sqrt[]{5}}$ = $\lim_{x \to5} \frac{(\sqrt[]{x}-\sqrt[]{5}).(\sqrt[]{x}+\sqrt[]{5})}{\sqrt[]{x}-\sqrt[]{5}}$
= $\lim_{x \to5}( x+5)$ = 5 + 5 = 10
b, $\lim_{x \to0} \frac{(1+x)³-1}{x}$ = $\lim_{x \to0} \frac{[(1+x)-1].[(1+x)²+(1+x).1+1²]}{x}$
= $\lim_{x \to0} \frac{x.[(1+x)²+x+2]}{x}$ = $\lim_{x \to0}[(1+x)²+x+2] $ = (1+0)² + 0 + 2 = 3
c, $\lim_{x\to 1} \frac{\sqrt[]{x}-1}{\sqrt[]{x+3}-2}$ = $\lim_{x\to 1} \frac{(\sqrt[]{x}-1).(\sqrt[]{x+3}+2)}{(\sqrt[]{x+3}-2).(\sqrt[]{x+3}+2)}$
= $\lim_{x\to 1} \frac{(\sqrt[]{x}-1).(\sqrt[]{x+3}+2)}{x+3-2²}$ = $\lim_{x\to 1} \frac{(\sqrt[]{x}-1).(\sqrt[]{x+3}+2)}{x-1}$
= $\lim_{x\to 1} \frac{(\sqrt[]{x}-1).(\sqrt[]{x+3}+2)}{y}$ = $\lim_{x\to 1} \frac{(\sqrt[]{x}-1).(\sqrt[]{x+3}+2)}{(\sqrt[]{x}-1).(\sqrt[]{x}+1)}$
= $\lim_{x\to 1} \frac{\sqrt[]{x+3}+2}{\sqrt[]{x}+1}$ = $\frac{\sqrt[]{1+3}+2}{\sqrt[]{1}+1}$ = 2
d, $\lim_{x \to0} \frac{1}{x²}.(\frac{1}{x²+1}-1)$ = $\lim_{x \to0} \frac{1}{x²}.[\frac{1-(x²+1)}{x²+1}]$
= $\lim_{x \to0} \frac{1}{x²}.\frac{-x²}{x²+1}$ = $\lim_{x \to0} \frac{-1}{x²+1}$ = $\frac{-1}{0²+1}$ = -1
