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How does one find the following limits using the most basic limit properties (and also the limit definition for $ e $)?

a)$\lim_{x\rightarrow +\infty}\large{\frac{\sqrt[3]{x^2\sqrt{x^2+1}}+3x}{x}} $

b)$\lim_{x\rightarrow 0^+} \large{(2-e^{\arcsin^2\sqrt{x}})^{\frac{3}{x}}}$

c) $\lim_{x\rightarrow -\pi}\large{\frac {\sin(4x)}{x^2+\pi x}}$

Update: In the original version the limit in b) was with $x \to 0$. In this case since $\sqrt{x}$ is defined only for $x \geq 0$, the limit does not exist. So changed the limit to $x \to 0^+$.

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5 Answers 5

For $(c)$ put $\displaystyle x+\pi=y\implies \sin4x=\sin 4(y-\pi)=\sin(4y-4\pi)=\sin4y$

$\displaystyle\lim_{x\rightarrow -\pi}\frac {\sin(4x)}{x^2+\pi x}=\lim_{y\to0}\frac{\sin 4y}{y(y-\pi)}$ Use $\lim_{u\to0}\frac{\sin u}u=1$

For $(a),$ $$\lim_{x\rightarrow +\infty}\frac{\sqrt[3]{x^2\sqrt{x^2+1}}+3x}x =\lim_{x\rightarrow +\infty}\frac{\sqrt[3]{\sqrt{x^6+x^4}}+3x}x=\lim_{x\rightarrow +\infty}\frac{(x^6+x^4)^{\frac16}+3x}x $$

Putting $\frac1x=h$, $$=\lim_{h\to0}(1+h^2)^{\frac16}+3=\cdots$$

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For $a$ (since $x>0$)$$\frac{\sqrt[3]{x^2\sqrt{x^2+1}}+3x}{x}=\frac{\sqrt[3]{x^2\sqrt{x^2+1}}}{\sqrt[3]{x^3}}+3=\sqrt[3]{x^{-1}\sqrt{x^2+1}}+3=\sqrt[3]{\sqrt{x^{-2}}\sqrt{x^2+1}}+3=\sqrt[3]{\sqrt{1+x^{-2}}}+3$$

$$\lim_{x\to\infty}\sqrt[3]{\sqrt{1+x^{-2}}}+3=\sqrt[3]{\sqrt{1+0}}+3=4$$

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Thanks. Can you help me with b? –  Constantine Oct 21 '13 at 17:21

I will try the limit in b) as other parts of the question have been solved quite nicely.

Let us write $L = \lim_{x \to 0}(2 - e^{\arcsin^{2}\sqrt{x}})^{3/x}$

Then we have

$\displaystyle \begin{aligned} \log L &= \log\left\{\lim_{x \to 0}(2 - e^{\arcsin^{2}\sqrt{x}})^{3/x}\right\} \\ &= \lim_{x \to 0}\log\left\{(2 - e^{\arcsin^{2}\sqrt{x}})^{3/x}\right\} \text{ (because of continuity of log) }\\ &= \lim_{x \to 0}\frac{3}{x}\log(2 - e^{\arcsin^{2}\sqrt{x}})\\ &= 3\lim_{x \to 0}\frac{\log(1 + 1 - e^{\arcsin^{2}\sqrt{x}})}{1 - e^{\arcsin^{2}\sqrt{x}}}\cdot\frac{1 - e^{\arcsin^{2}\sqrt{x}}}{x}\\ &= 3\lim_{x \to 0} 1\cdot \frac{1 - e^{\arcsin^{2}\sqrt{x}}}{x}\text{ (see explanation point 1 below) }\\ &= -3 \lim_{x \to 0} \frac{e^{\arcsin^{2}\sqrt{x}} - 1}{x}\\ &= -3\lim_{x \to 0} \frac{e^{\arcsin^{2}\sqrt{x}} - 1}{{\arcsin^{2}\sqrt{x}}}\cdot\frac{{\arcsin^{2}\sqrt{x}}}{x}\\ &= -3\lim_{x \to 0}1\cdot\left(\frac{\arcsin\sqrt{x}}{\sqrt{x}}\right)^{2}\text{ (see explanation point 2 below) }\\ &= -3\cdot 1^{2}\text{ (see explanation point 3 below) }\\ &= -3\end{aligned}$

Hence desired limit $L$ is given by $L = e^{-3}$.

Explanation for some steps: For calculating limits we can use following simple results:

1) $\displaystyle \lim_{y \to 0}\frac{\log(1 + y)}{y} = 1$

In the above problem we have $y = 1 - e^{\arcsin^{2}\sqrt{x}}$ so that $y \to 0$ as $x \to 0$ and hence the logarithmic limit evaluates to $1$.

2) $\displaystyle \lim_{y \to 0}\frac{e^{y} - 1}{y} = 1$

In this problem $y = \arcsin^{2}\sqrt{x}$ so that $y \to 0$ as $x \to 0$ and the exponential limit is also $1$.

3) $\displaystyle \lim_{y \to 0}\frac{\sin y}{y} = 1,\,\,\lim_{y \to 0}\frac{\arcsin y}{y} = 1$

Here $y = \sqrt{x}$ so that the arcsin limit evaluates to $1$.

In most common problems on limits these three fomulas suffice. When they don't (especially in some tough ones) I use L'Hospital Rule and if L'Hospital fails or generates complicated expression then only I opt for series expansions.

Update: The way question has been put the limit does not exist. We should change question so that $x \to 0+$.

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$$\lim_{x\to0}\left(2-e^{(\arcsin\sqrt x)^2}\right)^{\frac3x}$$

$$=\left(\lim_{x\to0}\left(1+1-e^{(\arcsin\sqrt x)^2}\right)^{\frac1{1-e^{(\arcsin\sqrt x)^2}}}\right)^{\lim_{x\to0}\frac{3(1-e^{(\arcsin\sqrt x)^2})}x}$$

For the inner limit put $\displaystyle1-e^{(\arcsin\sqrt x)^2}=\frac1n$ as $x\to0,n\to\infty\implies \lim_{n\to\infty}(1+\frac1n)^n=e$

For the limit in the power $$\lim_{x\to0}\frac{3(1-e^{(\arcsin\sqrt x)^2})}x=-3\lim_{x\to0}\frac{e^{(\arcsin\sqrt x)^2}-1}{(\arcsin\sqrt x)^2}\cdot \lim_{x\to0}\left(\frac{\arcsin\sqrt x}{\sqrt x}\right)^2$$

Put $(\arcsin\sqrt x)^2=h$ for the first limit and $\arcsin\sqrt x=u$ for the second to find $$\lim_{x\to0}\frac{3(1-e^{(\arcsin\sqrt x)^2})}x=-3$$

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@Constantine, I have answered $(b)$ separately for its complexity –  lab bhattacharjee Oct 21 '13 at 18:44
    
The last step doesn't seem to help, you still have the indetermined $\frac {0}{0} $ form. –  Constantine Oct 21 '13 at 19:04
    
@Constantine, have you noticed $$\lim_{x\to0}\frac{e^{(\arcsin\sqrt x)^2}-1}{(\arcsin\sqrt x)^2}=1$$ and $$\lim_{x\to0}\frac{\arcsin\sqrt x}{\sqrt x}=1$$ –  lab bhattacharjee Oct 21 '13 at 19:06
    
I have noticed the second one, but the first one I don't understand. Where does it come from? –  Constantine Oct 21 '13 at 19:14
    
@Constantine, are you talking about math.stackexchange.com/questions/42679/…? –  lab bhattacharjee Oct 22 '13 at 4:28

Here is an approach. Notice that, when $x$ close to zero, then we have by Taylor series

$$ \arcsin^2\sqrt{x} \sim x. $$

So,

$$ {(2-e^{\arcsin^2\sqrt{x}})^{\frac{3}{x}}}\sim {(2-e^{x})^{\frac{3}{x}}}=e^{\frac{3}{x}\ln(2-e^x)} = e^{\frac{3}{x}(-x+O(x^2))}\longrightarrow_{x\to 0} e^{-3} $$

Note:

$$ \ln(2-e^{x})=-x-x^2+O(x^3). $$

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