Find the value of $$\lim \limits_{x\to 0} \frac{\tan\sqrt[3]x\ln(1+3x)}{(\tan^{-1}\sqrt{x})^2(e^{5\large \sqrt[3]x}-1)}$$ Applying L'Hospital's rule does not seem to simplify the expression.

  • 2
    $\begingroup$ Have you tried Taylor expansion? $\endgroup$
    – Git Gud
    Nov 2 '13 at 15:19
  • $\begingroup$ Apply power series for each term in the expression. I got a similar problem in my maths olympiad. $\endgroup$
    – GTX OC
    Nov 2 '13 at 15:34

Note that $$\lim \limits_{y\to 0}\frac{\tan y}{y}=\lim_{y\to 0}\frac{\tan^{-1} y}{y}=\lim_{y\to 0}\frac{\ln (1+y)}{y}=\lim_{y\to 0}\frac{e^y-1}{y}=1.$$ Therefore,

$$\lim \limits_{x\to 0^+} \frac{\tan\sqrt[3]x\cdot \ln(1+3x)}{\left(\tan^{-1}\sqrt{x}\right)^2\cdot(e^{5 \sqrt[3]x}-1)}=\lim_{x\to 0^+}\frac{\sqrt[3]x\cdot 3x}{(\sqrt{x})^2\cdot 5 \sqrt[3]{x}}=\frac{3}{5}.$$

Edit: Some comments below suggest that the answer above may not be detailed enough. Let me explain a little more about the gap between the two lines of equations above. The limit we are concerned about can be written as

$$\lim \limits_{x\to 0^+}\left(\frac{\tan\sqrt[3]x}{\sqrt[3]x}\cdot \frac{\ln(1+3x)}{3x}\cdot \left(\frac{\sqrt{x}}{\tan^{-1}\sqrt{x}}\right)^2\cdot\frac{5 \sqrt[3]x}{e^{5 \sqrt[3]x}-1}\cdot \frac{\sqrt[3]x\cdot 3x}{(\sqrt{x})^2\cdot 5 \sqrt[3]{x}}\right). $$ On the one hand, as $x\to 0^+$, $\sqrt[3]x$, $3x$ $\sqrt x$ and $5\sqrt[3]x$ all approach to $0$, so from the first displayed line we know that each limit of first four terms in the product above exists and equals $1$. On the other hand, the limit of the last term exists and equals $\dfrac{3}{5}$. As a result, the original limit exists and equals $\dfrac{3}{5}$.

  • $\begingroup$ I don't see how you get the first equality in the second line from the first line. $\endgroup$
    – Git Gud
    Nov 2 '13 at 15:36
  • $\begingroup$ @GitGud: Substitute $y=\sqrt[3]{x}$, $y=3x$, $y=\sqrt{x}$ and $y=5\sqrt[3]{x}$ respectively. $\endgroup$
    – user104254
    Nov 2 '13 at 15:39
  • $\begingroup$ I thought you were using taylor series implicitly, I just realised you aren't. (+1) $\endgroup$
    – Git Gud
    Nov 2 '13 at 15:43
  • $\begingroup$ It seems that behaviour of functions of arguments of $x$ is similar to the behaviour of arguments of $x$, as $x$ tends to zero, after observing the second line. Is it rigorous? $\endgroup$
    – Tejas
    Nov 2 '13 at 15:48
  • 1
    $\begingroup$ @TBrendle I think you're incurring on the falacy of not looking at $=$ as symmetric relation, but only in one direction. For instance $\lim \limits_{x\to 0}\left(\dfrac{\sin(x)}{x}\dfrac{1}{\cos(x)}\right)=\lim \limits_{x\to 0}\left(\dfrac{\sin (x)}{x}\right)\lim \limits_{x\to 0}\left(\dfrac 1{\cos(x)}\right)=1\cdot 1=1$ is perfectly rigorous simply because you can read the equalities backwards and symmetry of $=$ takes care it. $\endgroup$
    – Git Gud
    Nov 2 '13 at 16:24

If you change variable (x = y^3) and expand as a Taylor series at y=0, the first terms are 3/5 - 3 y / 2 + 29 y^2 / 20 +...


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