Question

${lim}_{x\to 2}\frac{x^7-128}{x^5-32}$

Answer 1

${lim}_{x⟶2}\frac{x^7-128}{x^5-32}={lim}_{x⟶2}\frac{x^7-2^7}{x^5-2^5}={lim}_{x⟶2}\frac{\frac{x^7-2^7}{x-2}}{\frac{x^5-2^5}{x-2}}=\frac{{lim}_{x⟶2}\frac{x^7-2^7}{x-2}}{{lim}_{x⟶2}\frac{x^5-2^5}{x-2}}=\frac{7.2^6}{5.2^4}=\frac{7}{5}\times 2^2=\frac{7}{5}\times 4=\frac{28}{5}$