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I've managed to solve it by rewriting the expression as $$\frac{1 - \frac{\sqrt{x^2 +x + 1}}{x e^{\frac{1}{x}}} }{ \frac{1}{x e^{\frac{1}{x}}} }$$

then applying L'Hospital's rule.

This took up one whole page and was very hairy, even after substituting $t = \sqrt{x^2+x+1}$. I'm wondering if there's a simpler way. A friend suggested substituting $e = (1+\frac{1}{x})^x$, but that's a bit suspicious.

In both cases, the answer is $\frac{1}{2}$, as confirmed by my computer.

user2244484
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3 Answers3

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HINT:

Note that can write

$$\begin{align}xe^{1/x}=x\left(1+\frac1x+O\left(\frac1{x^2}\right)\right)\end{align}$$

and

$$\begin{align} \sqrt{x^2+x+1}&=x\left(1+\frac1x+\frac{1}{x^2}\right)^{1/2}\\\\ &=x\left(1+\frac{1}{2x}+O\left(\frac1{x^2}\right)\right) \end{align}$$

Mark Viola
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  • I'm afraid I don't follow. What is this $O\left(\frac{1}{x^2}\right)$ notation? Is it in some way related to "big oh" in computer science? I've never seen a similar approach to solving limits. – user2244484 Nov 27 '16 at 11:33
  • The notation is called "big O." Here, $f(x) =O(g(x))$ as $x\to \infty$ if and only if there exists a number $M$ such that for $x$ sufficiently large, $|f(x)|\le M|g(x)|$. – Mark Viola Nov 27 '16 at 14:37
3

Using the standard limit $$ \lim_{t\to 0}\frac{e^t-1}{t}=1 $$ we have $$ 1=\lim_{x\to\infty}\frac{e^{1/x}-1}{1/x}=\lim_{x\to\infty}(xe^{1/x}-x). $$ Now rewrite your limit as $$ \lim_{x\to\infty}(xe^{1/x}-x+x-\sqrt{x^2+x+1})=1+\lim_{x\to\infty}(x-\sqrt{x^2+x+1}). $$ To calculate the last limit we rewrite again \begin{align} &\lim_{x\to\infty}(x-\sqrt{x^2+x+1})=\lim_{x\to\infty}\frac{(x-\sqrt{x^2+x+1})(x+\sqrt{x^2+x+1})}{x+\sqrt{x^2+x+1}}=\\ &=\lim_{x\to\infty}\frac{x^2-(x^2+x+1)}{x+\sqrt{x^2+x+1}}= \lim_{x\to\infty}\frac{-x-1}{x+\sqrt{x^2+x+1}}=\lim_{x\to\infty}\frac{-1-\frac1x}{1+\sqrt{1+\frac1x+\frac{1}{x^2}}}=-\frac{1}{2}. \end{align}

A.Γ.
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0

Multiplying by $\displaystyle\frac{x e^{\frac{1}{x}} + \sqrt{x^2+x+1}}{x e^{\frac{1}{x}} + \sqrt{x^2+x+1}}$ we have $$\lim_{x\to\infty}\left(x e^{\frac{1}{x}} - \sqrt{x^2+x+1} \right)=\lim_{x\to\infty}\frac{x^2 (e^{\frac{2}{x}} -1)-x-1 }{x e^{\frac{1}{x}} + \sqrt{x^2+x+1} }=\lim_{x\to\infty}\frac{x (e^{\frac{2}{x}} -1)-1-1/x }{ e^{\frac{1}{x}} + \sqrt{1+1/x+1/x^2} }.$$

Now $\displaystyle\lim_{x\to\infty}x(e^{\frac{2}{x}} -1)=\lim_{x\to\infty}\frac{e^{\frac{2}{x}} -1}{1/x}=\lim_{x\to\infty}\frac{e^{\frac{2}{x}}(-2/x^2)}{-1/x^2}=2$.

Thus the final limit is: $\frac{2-1}{1+\sqrt{1}}=1/2$.

Ruzayqat
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