Prove that:lim
without L'Hopital's rule.
I don't know if this is possible.
WolframAlpha agrees with this limit.
There's a similar limit without L'Hopital's rule \lim_{x\to 0^+}\frac{\ln(\sin x)}{\ln (x)}=1, which is easier to prove and some proofs can be seen in this question.
In \lim_{x\to 0^+}\frac{\ln(\sin x)}{\ln (x)}=1 you can see x\to 0^+, which is important, but in this case \frac{\ln(\cos x)}{\ln\left(1-\frac{x^2}{2}\right)} is an even function and we can use x\to 0.
I've tried some of the methods given in the linked question and none of them worked.
We could use \lim_{x\to 0}\frac{\cos x}{1-\frac{x^2}{2}}=1.
Edit: I've noticed that also
\lim_{x\to 0}\frac{\cos (\cos x)}{\cos\left(1-\frac{x^2}{2}\right)}=1
And also \lim_{x\to 0}\frac{\sin (\cos x)}{\sin\left(1-\frac{x^2}{2}\right)}=1
And also
\lim_{x\to 0}\frac{\tan (\cos x)}{\tan\left(1-\frac{x^2}{2}\right)}=1
And also \lim_{x\to 0}\frac{\arcsin(\cos x)}{\arcsin\left(1-\frac{x^2}{2}\right)}=1
Etc.
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