limitation
- Thread starter J2YoYo
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medicalphysicsguy
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L'Hopitals' rule
According to L'Hopital's rule, the limit as x->0 of f(x)/g(x) is the same as the limit as x->0 of f'(x)/g'(x) . So your next step is to find those derivatives and see if your answer is not indeterminate any more. If it's still indeterminate, but getting less complicated, a second derivative might bring the answer.
If you show your work you can be helped more.
According to L'Hopital's rule, the limit as x->0 of f(x)/g(x) is the same as the limit as x->0 of f'(x)/g'(x) . So your next step is to find those derivatives and see if your answer is not indeterminate any more. If it's still indeterminate, but getting less complicated, a second derivative might bring the answer.
If you show your work you can be helped more.
Sorry,the lecturer havent teach this stuff but his assignment want us to solve this Q. so i hope to see a full answer and step here.
pka
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You should know that \(\displaystyle \lim _{t \to 0} \dfrac{{\sin (t)}}{t} = 1\).
Now use the above: \(\displaystyle \dfrac{{\tan \left( {3x^2 } \right)}}{{x^2 }} = \dfrac{3}{{\cos \left( {3x^2 } \right)}}\dfrac{{\sin \left( {3x^2 } \right)}}{{3x^2 }}\)
pka
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That is correct.