I am computing the following limit
lim
where a is a parameter. the case a= 0 has been resolved. But yet, for the case a\neq 0 I want to compute it without using L'Hopital rule.
Answer
If you know how to calculate the limit of \frac{x^3}{x-\sin x} without L'Hospital, then the rest follows from the mean value theorem for integrals (if we may use it - it is quite similar to L'Hospital/Bernoulli's rule)
\int_0^x\frac{t^2}{x^3\sqrt{a+t}}\,dt=\left[u=\frac{t}{x}\right]=\int_0^1\frac{u^2}{\sqrt{a+xu}}\,du=\frac{1}{\sqrt{a+x\xi}}\int_0^1u^2\,du=\frac{1}{\sqrt{a+x\xi}}\cdot\frac13\to\frac{1}{3\sqrt{a}}.
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