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Choose correct options , more than one may be correct .

Let f be the function defined by $f(x)=e^{\sqrt{x+\sqrt{1+x}}-\sqrt{x}}$ we've:

(1)

$\lim_{x\to\infty}f(x)=+\infty$

(2)

$\lim_{x\to\infty}f(x)=0$

(3)

$\lim_{x\to\infty}f(x)=e^{\dfrac{1}{2}}$

(4)

$\lim_{x\to\infty}f(x)=\sqrt{e}$

her graph enter image description here

I think the correct answer is (3) Indeed :

$$ \begin{align*} &\lim_{x\to +\infty} e^{\sqrt{x+\sqrt{1+x}}-\sqrt{x}}\\ =& \lim_{x\to +\infty} e^{\dfrac{\sqrt{1+x}}{\sqrt{x+\sqrt{1+x}}+\sqrt{x}}}\\ =&\lim_{x\to +\infty}e^{ \dfrac{\sqrt{x}\sqrt{\dfrac{1}{x}+1}}{\sqrt{x}(1+\sqrt{1+\dfrac{ \sqrt{1+x}}{x}})}}=\lim_{x\to +\infty}e^{ \dfrac{\sqrt{\dfrac{1}{x}+1}}{(1+\sqrt{1+\dfrac{ \sqrt{1+X}}{x}})}}=e^{\dfrac{1}{2}} \end{align*} $$

because of $\lim_{x\to +\infty}\dfrac{ \sqrt{1+x}}{x}=\lim_{x\to +\infty}\dfrac{ 1}{x}=0$

  • I wonder if there are some other short ways to calculate that limit ?
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The answer is (3) and (4): $\sqrt{e}$. When dealing with exponent like this one, it is always good to remember that if you let $$L = \lim_{x \to \infty} \exp\Big(\sqrt{x + \sqrt{1+x}} - \sqrt{x}\Big),$$ then $$\begin{align*} \ln L &= \ln \lim_{x \to \infty} \exp\Big(\sqrt{x + \sqrt{1+x}} - \sqrt{x}\Big) = \lim_{x \to \infty} \ln \exp\Big(\sqrt{x + \sqrt{1+x}} - \sqrt{x}\Big) \end{align*},$$ which is just $$ \ln L = \lim_{x \to \infty} \sqrt{x + \sqrt{1+x}} - \sqrt{x}.$$

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    $\begingroup$ Just to point out, $\sqrt{e}=e^{1/2}$ $\endgroup$ – Guy Apr 6 '14 at 15:55
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    $\begingroup$ Wow, can't believe I missed that, what a shame. It's early morning here... Anyways, I think the method described above could save him some time. $\endgroup$ – Hubble Apr 6 '14 at 15:58
  • $\begingroup$ I agree. +1${}{}$ $\endgroup$ – Guy Apr 6 '14 at 16:04
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I think the easiest way is to use Taylor series. We have that $$\sqrt{x+a}=\sqrt{x}+{1\over 2}x^{-{1\over 2}}a-...$$ Here $a=\sqrt{x+1}$, so we get $$\lim{\sqrt{x+a}-\sqrt{x}}=\frac{1}{2}$$ because the other terms in the Taylor series tend to 0.

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    $\begingroup$ You would be hard pressed to know taylor series without knowing such elementary limits. Just my opinion. $\endgroup$ – Guy Apr 6 '14 at 16:05
  • $\begingroup$ @ Sabyasachi true but we want to see most short method $\endgroup$ – Adam Apr 6 '14 at 16:11

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