Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 2

Write $R(x)=(x^2+3)(2x^2-1)/(3x^4-x^3-1)$. This rational function is continuous at both finite points because its denominator is nonzero there. Substitution gives the first two limits; at infinity divide numerator and denominator by $x^4$ and use their leading coefficients. Therefore

For fixed real $a$, the base is positive for all sufficiently large positive $x$. Take its logarithm and use the Taylor expansion $\ln (1-u)=-u+O(u^2)$:Continuity of the exponential function then gives $\boxed{\underset{x\to \infty }{\lim }(1-a/x{)}^x=e^{-a}}$. This includes $a=0$.

The numerator and denominator both vanish at zero. The fundamental theorem of calculus differentiates the numerator, while the denominator derivative is nonzero near zero. L'Hôpital's rule therefore gives

Both numerator and denominator have a double zero at one. Applying L'Hôpital's rule twice givesAlternatively $x^3-3x+2=(x-1{)}^2(x+2)$ and $1-\sin (\pi x/2)\sim {\pi }^2(x-1{)}^2/8$, giving the same result from a Taylor expansion.