\(\lim_{x\rightarrow 0}\)\((x+1)^5-\frac{1}{x}\) Put x+1 = y so that y→1 as x → 0. Accordingly, \(\lim_{x\rightarrow 0}\)\((x+1)^5-\frac{1}{x}\) = \(\lim_{y\rightarrow 1}\)\(\frac{y^{5-1}}{y-1}\) = \(\lim_{y\rightarrow 1}\)\(\frac{y^{5-1}}{y-1}\) =5 ∴\(\lim_{x\rightarrow 0}\)\((x+1)^5-\frac{1}{x}\) = 5