Question

Lim x-->1 [sin (x-1)] / [(x^5)-1]

Answer 1

`x^5-1=(x-1)(x^4+x^3+x^2+x+1)`

so you can write


`\displaystyle\lim_(x\to1)(\sin(x-1))/((x-1)(x^4+x^3+x^2+x+1))=\left(\lim_(x\to1)(\sin(x-1))/(x-1)\right)\left(\lim_(x\to1)\frac1{x^4+x^3+x^2+x+1}\right)=\frac15`

= = = = = = = = = = = = = = = = =

Alternate method: By L'Hopital's rule, you have


`\displaystyle\lim_(x\to1)(\sin(x-1))/(x^5-1)=\lim_(x\to1)(\cos(x-1))/(5x^4)=\frac{\cos0}5=\frac15`

Answer 2

`\bf \lim\limits_(x\to 1)\ \cfrac{sin(x-1)}{x^5-1}\\\\ -----------------------------\\\\ u=x-1\qquad 1=1^5\\\\ -----------------------------\\\\ \cfrac{sin(x-1)}{x^5-1^5}\implies \cfrac{sin(x-1)}{(x-1)(x^4+x^3+x^2+x+1)} \\\\\\ \cfrac{sin(u)}{u(x^4+x^3+x^2+x+1)}\implies \cfrac{sin(u)}{u}\cdot \cfrac{1}{(x^4+x^3+x^2+x+1)} \\\\\\ \lim\limits_(u\to 0)\ \cfrac{sin(u)}{u}\quad \cdot \quad \lim\limits_(x\to 1)\cfrac{1}{(x^4+x^3+x^2+x+1)} \\\\\\ 1\cdot \cfrac{1}{1+1+1+1+1}\implies \cfrac{1}{5}`