Question

(1) check whether the function f given by f(x) = 1/(2x-4)^2 for all x belonging to ]-2,2[ is continuous in the interval ]-2,2[. Is it bounded? Justify your answer.

(2) find the values of p and q so that lim { p sinx + x(1-cosx)}/5x^3 as x tends to infinity is 1/6

Answer 1

(1) Consider a function $f(x)=\frac{1}{(2x-4)^2}, \ x \in ]-2,2[$ . The gived function is continuous on interval $]-2,2[$ as a ratio of two continuous functions on $]-2,2[$.

The function $f(x)$ is unbounded in the interval $]-2,2[$ , because there exists a sequence $x_n=2-\frac1n, \ n \in \mathbb{N}$, in interval $]-2,2[$ , such that $f(x_n)=\frac{n^2}{4} \to \infty$ for $n \to \infty$.

(2) Let consider a function $g(x)=\frac{p\sin x+x(1-\cos x)}{5x^3}=\frac{p\sin x}{5x^3}+\frac{1-\cos x}{5x^2}.$ Since $p\sin x$ and $1-\cos x$ are bounded functions in $\mathbb{R}$, then $\lim\limits_{x \to \infty}\frac{p\sin x}{5x^3}=0$, $\lim\limits_{x \to \infty} \frac{1-\cos x}{5x^2}=0$. Therefore $\lim\limits_{x \to \infty} \frac{p\sin x+x(1-\cos x)}{5x^3}=0$ for every values of $p$ and $q$. Consequently, there is no values of parameters $p$ and $q$ for which $\lim\limits_{x \to \infty} \frac{p\sin x+x(1-\cos x)}{5x^3}=\frac{1}{6}.$