Question

${lim}_{x\to 1}{\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}^{\frac{1-\cos (x-1)}{(x-1{)}^2}}$

Answer 1

Let $y={lim}_{x\to 1}{\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}^{\frac{1-\cos (x-1)}{(x-1{)}^2}}$
Taking log on both sides, we gets
$logy=\frac{1-cos(x-1)}{(x-1{)}^2}log\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}y=e^{\frac{1-cos(x-1)}{(x-1{)}^2}log\left\{\frac{x^2+2x^2+x+1}{x^2+2x+3}\right\}}\therefore {\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}^{\frac{1-\cos (x-1)}{(x-1{)}^2}}=e^{\frac{1-cos(x-1)}{(x-1{)}^2}log\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}$
taking limit $x\to \infty$ we get
${lim}_{x\to \infty }{\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}^{\frac{1-cos(x-1)}{(x-1{)}^2}}=li{m}_{x\to \infty }{e}^{\frac{1-\cos (x-1)}{(x-1{)}^2}log\left\{\frac{x^3+2x^2+x+1}{x^2+2x+3}\right\}}$