Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1 / 3F (Analysis I)

For $0<x<\pi /2$, the usual unit-circle area comparison, with angles measured in radians, gives $\sin x\le x\le \tan x$. HenceThe cosine is a continuous function at zero, so the squeeze theorem yields the right-hand limit of a function $1$. The quotient $\sin x/x$ is even, so the left-hand limit of a function is the same. Thus the limit is $1$.

For $x$ sufficiently close to zero, $1+x>0$. Apply the mean value theorem to the logarithm between $1$ and $1+x$: there is a point ${\xi }_x$ between these endpoints withAs $x\to 0$ from either side, ${\xi }_x\to 1$, so this quotient tends to $1$. The exponential function is continuous, and thereforeThe hypotheses of the mean value theorem hold because $\log t$ is continuous and differentiable for $t>0$, with derivative $1/t$.

For $x>0$, the exponent $x/(1+x)$ lies in $(0,1)$, and $0\le {\cos }^{4}x\le 1$. ConsequentlyThe last inequality uses the positive-term power series $e^x={\sum }_{k\ge 0}{x}^k/k!\ge x^2/2$ for $x>0$. The final bound tends to zero, so the squeeze theorem gives limit $0$.