How can I calculate $\lim_{x \to \infty}((x^3-x^2+\frac{x}{2})e^{\frac{1}{x}}-\sqrt{x^6+1})$?

Question

Given: $$\lim_{x \to \infty}((x^3-x^2+\frac{x}{2})e^{\frac{1}{x}}-\sqrt{x^6+1})$$

How can I calculate this limit? I tried to calculate the limit by use of Taylor for $e^{\frac{1}{x}}$ but I don't have idea how to calculate it.

Answer 1

Hint:

Factor out $x^6$ in both terms to obtain a second factor in $u=\dfrac1x$, and expand $\sqrt{1+\dfrac1{x^6}}$ with the first terms of the binomial series.

Answer 2

I tried to calculate the limit by use of Taylor for $e^{\frac1x}$ but I don't have idea how to calculate it.

Hint. If one knows that, as $u \to 0$, $$ e^u=1+u+\frac {u^2}2+\frac{u^3}6+O(u^4) $$ then, as $x \to \infty$, one gets $$ e^{\large\frac1x}=1+\frac1x+\frac 1{2x^2}+\frac 1{6x^3}+O\left(\frac1{x^4} \right). $$ By multiplying $$ \left(x^3-x^2+\frac{x}{2}\right)e^{\large\frac{1}{x}} $$ one obtains a polynomial in terms of $\frac 1x$.

Similarly, as $x \to \infty$, by using a taylor expansion, one has $$ \sqrt{x^6+1}=x^3\cdot \sqrt{1+\frac 1{x^6}}=x^3\cdot\left(1+\frac 1{2x^6}+ O\left(\frac1{x^{12}} \right)\right) $$ which might help to conclude.

Answer 3

We have, using the fact that $1/x\to 0$ when $x\to \infty$, $$ e^{1/x} = 1+ \frac{1}{x}+ \frac{1}{2x^2}+ \frac{1}{6x^3} + o\left(\frac{1}{x^3}\right) $$ when $x\to \infty$. This will help you deal with the term $e^{1/x}$.

Similarly, to deal with the $\sqrt{x^6+1}$, we can factor and use the same trick: $$ \sqrt{x^6+1} = x^3\sqrt{1+\frac{1}{x^6}} = x^3\left(1+\frac{1}{2x^6}+o\left(\frac{1}{x^6}\right)\right) = x^3+o(1) $$ when $x\to\infty$.

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Now, $$\begin{align} \left(x^3-x^2+\frac{x}{2}\right)e^{\frac{1}{x}}-\sqrt{x^6+1} &= \left(x^3-x^2+\frac{x}{2}\right)e^{\frac{1}{x}}-x^3\sqrt{1+1/x^6}\\ &= \left(x^3-x^2+\frac{x}{2}\right)e^{\frac{1}{x}}-x^3\left(1+\frac{1}{2x^6}+o\left(\frac{1}{x^6}\right)\right)\\ &= \left(x^3-x^2+\frac{x}{2}\right)\left(1+\frac{1}{x}+ \frac{1}{2x^2}+ \frac{1}{6x^3}+o\left(\frac{1}{x^3}\right)\right)-\left(x^3+o(1)\right)\\ &= \left(x^3-x^2+\frac{x}{2}\right)\left(1+\frac{1}{x}+ \frac{1}{2x^2}+ \frac{1}{6x^3}+o\left(\frac{1}{x^3}\right)\right)-\left(x^3+o(1)\right)\\ &=\left(x^3 +x^2 + \frac{x}{2}+ \frac{1}{6}+o(1)\right)-\left(x^2+x+\frac{1}{2}+o(1)\right)+\left(\frac{x}{2}+\frac{1}{2}+o(1)\right)-(x^3+o(1))\\ &=\frac{1}{6}-\frac{1}{2}+\frac{1}{2}+o(1)) = \frac{1}{6}+o(1)\\ &\xrightarrow[x\to\infty]{} \boxed{\frac{1}{6}} \end{align}$$

Answer 4

Put $x=1/t$ so that $t\to 0^{+}$ and the expression under limit is transformed into $$\frac{(1-t+t^2/2)e^{t}-\sqrt{1+t^6}} {t^3} $$ Multiplying numerator and denominator by $(1-t+t^2/2)e^t+\sqrt{1+t^6}$ and noting that this factor tends to $2$ we can see that the desired limit is equal to the limit of the expression $$\frac {(1-t+t^2/2)^2e^{2t}-1-t^6}{2t^3}$$ which is same as the limit of $$\frac{(1-t+t^2/2)^2e^{2t}-1}{2t^3}$$ which is same as that of $$\frac{(1-t+t^2/2)^2-e^{-2t}}{2t^3}$$ Factoring the numerator via $a^2-b^2=(a-b)(a+b) $ and noting that $a+b\to 2$ it is easy to see that the desired limit is equal to the limit of $$\frac{1-t+t^2/2-e^{-t}}{t^3}$$ Now the Taylor series for $e^{-t} $ gives the answer $1/6$ immediately. I always avoid multiplication, division and composition of Taylor series unless absolutely necessary. That way the problem of figuring out the number of terms needed in the Taylor series is trivialized.