This Limit problem seems too simple....

[vela]

The Calc 1 approach to this problem is probably to multiply and divide by the conjugate:
\begin{align*}
\left(x^3 +x^2 +\frac{x}{2}\right)e^{\frac{1}{x}} - \sqrt{x^6+1}
&= \left[\left(x^3 +x^2 +\frac{x}{2}\right)e^{\frac{1}{x}} - \sqrt{x^6+1}\right]\frac{\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}} + \sqrt{x^6+1}} \\
&= \frac{\left(x^3 +x^2 +\frac{x}{2}\right)^2e^{\frac{2}{x}} - (x^6+1)}{\left(x^3 +x^2 +\frac{x}{2}\right)e^{\frac{1}{x}} + \sqrt{x^6+1}} \\
\end{align*} It's fairly straightforward to analyze the problem from here with the usual Calc 1 methods.

 

[archaic]

Generally, you can't because cancellation may occur. If you can see that cancellation won't occur, then you can do the replacement. In this thread's problem, the leading order term is going to cancel, so it's not obvious you can simply discard the lower-order terms in the polynomial.

So how can generally foresee that it is not OK to substitute a function with its asymptotic equivalent?

 

[fresh_42]

So how can generally foresee that it is not OK to substitute a function with its asymptotic equivalent?

It is ok. You simply have to consider all factors, quotients, sums and differences to determine what asymptotic really is!