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Trigonometric limit

Yankel

Active member
Jan 27, 2012
398
Hello,

I need some help with this limit, I have no clue how to do this with the next constraint: No use of L'hopital rule...

Thank you !

[tex]\lim_{x\to1}\frac{sin(x^{3}-1)}{x-1}[/tex]
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello,

I need some help with this limit, I have no clue how to do this with the next constraint: No use of L'hopital rule...

Thank you !

[tex]\lim_{x\to1}\frac{sin(x^{3}-1)}{x-1}[/tex]
Try multiplying numerator and denominator by $x^{2} + x + 1$...

Kind regards

$\chi$ $\sigma$
 

Yankel

Active member
Jan 27, 2012
398
Try multiplying numerator and denominator by $x^{2} + x + 1$...

Kind regards

$\chi$ $\sigma$
I thought about it, it gives me what's in the sine, however x->1 and not x->0, so I am still stuck...
 

chisigma

Well-known member
Feb 13, 2012
1,704
I thought about it, it gives me what's in the sine, however x->1 and not x->0, so I am still stuck...
$\displaystyle \lim_{x \rightarrow 1} \frac{\sin (x^{3}-1)}{x-1} = \lim_{x \rightarrow 1} \frac{\sin (x^{3}-1)}{x^{3}-1}\ (x^{2} + x + 1)$

... and now what is $\displaystyle \lim_{x \rightarrow 1} \frac{\sin (x^{3}-1)}{x^{3}-1}$?...

... and what is $\displaystyle \lim_{x \rightarrow 1} x^{2} + x + 1$?...

Kind regards

$\chi$ $\sigma$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Also, L'Hôpital's rule allows us to write:

$\displaystyle \lim_{x\to1}\frac{\sin(x^3-1)}{x-1}=3\lim_{x\to1}x^2\cos(x^3-1)$