limits

[291]

lim x -->0 sin (x^2)/ 3x^2

 

[daon2]

lim x -->0 sin (x^2)/ 3x^2

Make the substitution $\displaystyle u=x^2$ and note that $\displaystyle x\to 0 \iff u\to 0$

$\displaystyle \displaystyle\lim_{x\to 0}\frac{\sin(x^2)}{3x^2} = \frac{1}{3}\lim_{u\to 0}\cdot \frac{\sin(u)}{u}$

 

[lookagain]

lim x -->0 sin(x^2)/(3x^2) . . . . Make sure you place grouping symbols around this denominator.

Alternative using L'Hopital's Rule:


$\displaystyle \displaystyle\lim_{x\to 0} \dfrac{[cos(x^2)](2x)}{6x} \ =$


$\displaystyle \displaystyle\lim_{x\to 0} \dfrac{cos(x^2)}{3} \ = \ ?$

 

[biffboy]

lim x -->0 sin (x^2)/ 3x^2

=1/3limx->0 sinx^2/x^2 Hospital rule =1/3(2xcosx^2)/2x=limx->0 1/3(cosx^2)= 1/3

 

[lookagain]

=1/3limx->0 sinx^2/x^2 Hospital rule =1/3(2xcosx^2)/2x=limx->0 1/3(cosx^2)= 1/3

I already posted this solution using the same method right before yours, biffboy.