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limits
- Thread starter 291
- Start date
lim x -->0 sin (x^2)/ 3x^2
Make the substitution \(\displaystyle u=x^2\) and note that \(\displaystyle x\to 0 \iff u\to 0\)
\(\displaystyle \displaystyle\lim_{x\to 0}\frac{\sin(x^2)}{3x^2} = \frac{1}{3}\lim_{u\to 0}\cdot \frac{\sin(u)}{u}\)
lim x -->0 sin(x^2)/(3x^2) . . . . Make sure you place grouping symbols around this denominator.
Alternative using L'Hopital's Rule:
\(\displaystyle \displaystyle\lim_{x\to 0} \dfrac{[cos(x^2)](2x)}{6x} \ = \)
\(\displaystyle \displaystyle\lim_{x\to 0} \dfrac{cos(x^2)}{3} \ = \ ?\)
Last edited:
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
=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.