Respuesta :

[tex]\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}[/tex]
LammettHash
[tex]x^5-1=(x-1)(x^4+x^3+x^2+x+1)[/tex]

so you can write

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

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

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

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

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