1
$\begingroup$

$\displaystyle\lim_{x\to0} (\sqrt x ~(\ln(x))^{2015})$

I used the L'hospital Rule. However I have to calculate a derivate of super high order. I think there is a formula for calculating the nth derivative of $\ln(x)$, or maybe I can calculate it from Taylor's series. As for $ \frac{1}{\sqrt x} $, I'm uncertain about how it should be laid out.

Is there a better method?

|cite|improve this question
$\endgroup$
  • $\begingroup$ Is $\ln^{2015}(x)$ $(\ln(x))^{2015}$ or the natural logarithm iterated 2015 times? $\endgroup$ – Michael McGovern Dec 18 '16 at 5:23
  • $\begingroup$ this is a greatest question..........-11111111111111 $\ldots \infty$ $\endgroup$ – Bhaskara-III Dec 18 '16 at 5:34
  • $\begingroup$ As the box above stresses, it is important, when having others solve your homework/exercises, that you (provide yourself with and) write down an attempt of solution and your knowledge of the subject. For instance, this problem could be approached in, say, $4$ ways, but you might not have the knowledge to understand one of them. Fact is that only you can provide that information, which ultimately saves people's time. $\endgroup$ – user228113 Dec 18 '16 at 6:57
  • $\begingroup$ Hint: Taylor expand ln(x) and disregard higher order terms $\endgroup$ – infinitylord Dec 18 '16 at 7:05
  • $\begingroup$ @infinitylord You cannot use Taylor expansion to estiamte effectively $\ln x$ in a neighbourhood of $0$, because $\lim\limits_{x\to 0^+}\ln x=-\infty$. $\endgroup$ – user228113 Dec 18 '16 at 16:03
2
$\begingroup$

Put $x=1/t$ and use the inequality $$\log t\leq t - 1$$ to show more generally that $$\lim_{t\to\infty} \frac{(\log t) ^{a}} {t^{b}} = 0$$ for any positive numbers $a, b$. Your case corresponds to $a=2015,b=1/2$.

Thus let $c=b/a$ and let $d$ be any number such that $0<d<c$. We have for $t>1$ $$0<d\log t = \log t^{d} \leq t^{d} - 1<t^{d}$$ or $$0<\log t<\frac{t^{d}} {d} $$ and hence $$0<\frac{\log t} {t^{c}} <\frac{1}{dt^{c-d}}$$ By squeeze theorem $(\log t) /t^{c} \to 0$ as $t\to\infty$. Raising to positive power $a$ we get $(\log t) ^{a} /t^{b} \to 0$ as $t\to\infty$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I would like to add that after putting x = 1/t, the logarithm must be transformed further to the proper form described in your answer. $\endgroup$ – DudeLearningStuffs Dec 24 '16 at 10:22
  • $\begingroup$ $\log t$ is undefined for $t<0$, i.e., whenever $x<0$, so this analysis falls apart unless you are addressing right hand limit only... $\endgroup$ – adfriedman Dec 24 '16 at 10:55
  • $\begingroup$ @user153126: The question actually makes good sense only if the limit is $x\to 0^{+}$. So the question needs to be corrected. However there is a relaxed definition of limit where the notation $x\to c$ only requires that $c$ be a limit point of domain of the function under consideration and then I don't see any issue here. But thanks for pointing out. $\endgroup$ – Paramanand Singh Dec 24 '16 at 13:21
1
$\begingroup$

In general if $k\in \mathbb N$ and $\lim_{x\to A}f(x)=L$ then $\lim_{x\to A}f(x)^k=L^k.$ This includes the cases $A=\pm \infty.$ This also includes one-sided limits when $A\in \mathbb R.$ E.g. if $x$ is restricted to $(A,\infty)$ we can just define $f(x)=L$ for $x\leq A$ if we want a double-sided limit.

For $x>0$ let $x=y^2$ with $y>0.$ Let $ k=2015.$ Then for $x>0$ we have $$( \sqrt x \cdot \log x)^{2015}=(y\log (y^2))^k=(2y\log y)^k.$$ If you know $\lim_{y\to 0+}y\log y=0$ you are done. If you didn't already know that, apply l'Hopital to $f(x)/g(x)$ with $f(x)=\log x$ and $g(x)=1/x.$

|cite|improve this answer
$\endgroup$
0
$\begingroup$

Writing it in the form $\lim_{x \to 0} \frac{ln^{2015}(x) }{\frac{1}{\sqrt{x}}}$ and applying L'Hopital's rule should lead to the answer: $0^-$.

See http://mathworld.wolfram.com/LHospitalsRule.html

|cite|improve this answer
$\endgroup$
  • $\begingroup$ @shivam It is true that L'Hopital can be used to solve the problem. However, it must be used repeatedly ($2015$ times, to be precise). To make your post an effective hint, it should at least cover this part of the procedure. $\endgroup$ – user228113 Dec 18 '16 at 6:46
0
$\begingroup$

First off, the limit does not exist as neither square root nor log function are defined for negative numbers (recall that a limit must exist from the left and right!). This is why the other posted answer is incorrect and, if anything, should emphasize the importance of not just applying l'Hopital's without checking the conditions first, they are not a minor technicality.

Suppose, on the other hand, that you want the right hand limit instead then we will use the fact that the exponential function grows faster than any polynomial, i.e., $\lim_{t\to\infty} \frac{t^n}{e^t} = 0$ for any $n$.

In the same way $e^t$ grows really fast, $1/e^t$ shrinks really quickly. To utilize this substitute $x=e^{-2t}$, then $$ \lim_{x \to 0^+} \sqrt{x} \log(x)^{2015} = \lim_{t\to \infty} e^{-t} (-2t)^{2015} = - 2^{2015} \lim_{t\to\infty} \frac{t^{2015}}{e^t} = 0 $$

Here we are implicitly using the fact that $e^{-2t}$ is one-to-one and continuous, and that $\sqrt{x}\log(x)^{2015}$ is right continuous at $0$.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.