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I'm asked to find the value of this limit:

$$\lim_{x\to\ 0^+} \arccos(\ln(1-2x)^{\frac{1}{4x}})$$

I am NOT allowed to use L'hopital's rule here.

My initial goal was try to first rewrite the limit as:

$$\arccos(\lim_{x\to\ 0^+}\ln(1-2x)^{\frac{1}{4x}})$$

The I was trying to use the fact that:

$$\lim_{x\to\ \infty }(1+\frac{1}{x})^x=\ e$$

So I made the substitution $x=\frac{-1}{2u}$ so therefore, $-2u=\frac{1}{x}$ so I get that as $x$ approaches $0$, $u$ approaches $\infty$. So overall, my limit becomes:

$$\arccos(\lim_{x\to\ 0^+}\ln(1+\frac{1}{u})^{-2u})^{\frac{1}{4}}$$

However, now, I become stuck. Is this method even the right approach? How would I even begin to go about solving this?

For reference, Here is the question:

enter image description here

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  • $\begingroup$ I think $\ln(1-2x)^\frac{1}{4x}$ means $\ln\Big((1-2x)^\frac{1}{4x}\Big)$ and not $\Big(\ln(1-2x)\Big)^\frac{1}{4x}$. $\endgroup$ – A.Γ. Oct 6 '17 at 6:35
  • $\begingroup$ Since $(1+u)^{\frac{1}{u}} \to e$ as $u \to 0$ thus $(1-2x)^{\frac{1}{4x}} \to e^{-\frac{1}{2}}.$ $\endgroup$ – M. Strochyk Oct 6 '17 at 6:38
  • $\begingroup$ Yeah I got it now. It was just a formatting error in the question. $\endgroup$ – Future Math person Oct 6 '17 at 7:11
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You can certainly pull out the arccosine, provided that $$ \lim_{x\to\ 0^+}\ln(1-2x)^{1/(4x)}\tag{*} $$ exists and belongs to $[-1,1]$.

Now I assume the exponential applies to $1-2x$ and so you can pull out also the logarithm, provided $$ \lim_{x\to\ 0^+}(1-2x)^{1/(4x)} $$ exists and is positive. Do the substitution $2x=1/t$ and the limit becomes $$ \lim_{t\to\infty}\biggl(\biggl(1-\frac{1}{t}\biggr)^{\!t}\biggr)^{\!1/2} $$ Can you go on from here?


The limit (*) can't be interpreted as $$ \lim_{x\to\ 0^+}\bigl(\ln(1-2x)\bigr)^{1/(4x)} $$ because the basis would be negative.

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  • $\begingroup$ Yeah. I already solved it just now. I suppose the question just wrote the problem strangely because I did not think the exponential applies to the inside of Ln. $\endgroup$ – Future Math person Oct 6 '17 at 8:49
  • $\begingroup$ @FutureMathperson One should avoid ambiguities, particularly in tests; unfortunately, it happens. $\endgroup$ – egreg Oct 6 '17 at 8:50

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