tag:blogger.com,1999:blog-9004452969563620551.post3799855340518374080..comments2017-06-21T11:55:27.964+05:30Comments on Paramanand's Math Notes: Two Problems not from IIT-JEEtesternoreply@blogger.comBlogger8125tag:blogger.com,1999:blog-9004452969563620551.post-6426415474667731672015-10-01T09:14:47.776+05:302015-10-01T09:14:47.776+05:30@Sid,
The Taylor series approach is very simple fo...@Sid,<br />The Taylor series approach is very simple for the first problem. In fact with a little amount of manipulation you can see that the desired limit is equal to $$\lim_{x \to 0}\frac{\sin x\sin^{-1}x - x^{2}}{x^{6}}$$ and using the Taylor series for both $\sin x$ and $\sin^{-1}x$ the answer is arrived at very easily.<br /><br />I consider Taylor series approach to be at a conceptually higher level than L'Hospital Rule (although the version of Taylor's Theorem used for evaluating limits can be proved using L'Hospital's Rule, see http://paramanands.blogspot.com/2013/11/teach-yourself-limits-in-8-hours-part-4.html) and hence prefer to avoid it for expository purposes.Paramanand Singhhttp://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-88060406226977515132015-09-30T22:46:55.288+05:302015-09-30T22:46:55.288+05:30I used taylor series and its manipulations to arri...I used taylor series and its manipulations to arrive at the answer in a fewer steps for the first question. The only question that would remain is the justification for the manipulation of the infinite series, which I'm sure can be given.Sidhttp://www.blogger.com/profile/10692920650937610398noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-48021818758732187522013-09-23T17:05:22.077+05:302013-09-23T17:05:22.077+05:30@Did,
Thanks once again for correcting me. Somehow...@Did,<br />Thanks once again for correcting me. Somehow I had this thinking that the lower limit of integration is dependent on $x$ whereas in reality it is dependent on $\varepsilon$paramanandhttp://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-52915088650592966622013-09-23T16:08:41.096+05:302013-09-23T16:08:41.096+05:30No. The lower limit is fixed and the upper limit g...No. The lower limit is fixed and the upper limit goes to infinity hence the “+constant terms” will be, surprise… a constant term, which disappears in the limit when $x\to\infty$ since it is divided by $e^x$.<br /><br />To sum up, you invented a problem where there is none.didhttp://www.blogger.com/profile/17176404801190299315noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-2814204808652735412013-03-20T10:01:23.140+05:302013-03-20T10:01:23.140+05:30Thanks Didier for providing a different viewpoint ...Thanks Didier for providing a different viewpoint based on integration. The argument is almost precise and rigorous except for the fact that during integration both the upper and lower limits must themselves be very large and then it will be difficult to reach the conclusion.Roughly the “+ constant terms” will actually not be constants but rather exponentials. The solution presented in the post does away with this problem.paramanandhttp://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-20600943097965079042013-03-20T10:00:22.566+05:302013-03-20T10:00:22.566+05:30For Problem 2, consider $g(x)=\mathrm e^xf(x)$, no...For Problem 2, consider $g(x)=\mathrm e^xf(x)$, note that $(a-\varepsilon)\mathrm e^x\leqslant g'(x)\leqslant(a+\varepsilon)\mathrm e^x$ for every $x$ large enough and integrate this. This yields $g(x)=(a\pm\varepsilon)\mathrm e^x+$ constant terms, for every $x$ large enough, hence $f(x)\to a$. QED.Didiernoreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-21984559166747585922013-03-20T09:56:36.493+05:302013-03-20T09:56:36.493+05:30Very informative post indeed.. being enrolled in I...Very informative post indeed.. being enrolled in IIT-JEE / AIEEE Exams Online Course: http://www.wiziq.com/course/7840-mathematics-physics-and-chemistry-for-iit-jee-aieee , I was looking for such articles online to assist me and your post will helped me a lot.Nancynoreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-34498589249876300642013-03-20T09:52:34.809+05:302013-03-20T09:52:34.809+05:30Readers should try the simpler (but similar) limit...Readers should try the simpler (but similar) limit problem:<br />$\displaystyle \lim_{x \to 0}\frac{x\tan(\tan x) - \tan^{2}x}{x^{6}}$paramanandhttp://www.blogger.com/profile/03855838138519730072noreply@blogger.com