Question: Hi I'm trying to manipulate; $\frac{\pi\cos(\pi x)}{\sqrt{\cos^2(\pi x)}}$; to; $\pi \sqrt{\cos^2(\pi x)} \sec(\pi x)$. ~ (From Wolfram but no steps).

When I try I get $\frac{\pi\cos(\pi x)}{\sqrt{\cos^2(\pi x)}} = \frac{1}{\sqrt{\cos^2(\pi x)}}\cdot \frac{\pi}{\sec(\pi x)} = \frac{\pi}{\sqrt{\cos^2(\pi x)}sec(\pi x)}$.

I'm unsure what else I can do.

~ Thanks.

Answer: 
Perhaps surprisingly, no. A standard counter-example is
$$f(x,y)=\left\{\begin{array}{ll} 0 & \text{ if } (x,y)=(0,0) \\ \dfrac{xy^2}{x^2+y^4} & \text{ if } (x,y)\not=(0,0) \end{array}\right.$$
$f$ is not continuous at $(0,0)$ since $f(y^2,y)=\dfrac{1}{2}$ for every $y\not=0$ while $f(0,0)=0$.
Let $f(x,0)=0$ for all $x.$ For $k\ne 0$ let $f(x,kx)=x/k.$ For $n\in \mathbb N$ let $(x\_n,y\_n)=(1/n, 1/n^2).$ Now for any $r>0$ we have $$0<y\_n<x\_n<r \land f(x\_n,y\_n)=1$$ for all sufficiently large $n\in \mathbb N.$

Equivalently let $f(x,y)=x^2/y$ for $y\ne 0,$ and $f(x,0)=0.$

Remarks:This is similar to the concept of uniform convergence: For functions $g,h$ from $D\subset \mathbb R$ to $\mathbb R,$ let $$\|g-h\|\_D=\min (1, \sup \{|g(x)-h(x)|:x\in D\}).$$ If $g(x)=\lim\_{n\to \infty}g\_n(x)$ for each $x\in D,$ for a sequence $(g\_n:D \to \mathbb R)\_{n\in \mathbb N},$ we say the convergence is uniform iff $$\lim\_{n\to \infty}\|g-g\_n\|\_D=0.$$ If the convergence is uniform and each $g\_n$ is continuous on $D,$ then $g$ will also be continuous on $D.$ But if the convergence is not uniform then $g$ may fail to be continuous on $D.$

For example let $D=[0,1]$ and $g\_n(x)=x^n.$ 

For this $Q$, with $D=\mathbb R$, it suffices that $f(x,x/n)=g\_n(x)$ for $n\in \mathbb N$, where $\lim\_{x\to 0}g\_n(x)=0$ for each $n,$ but that for any $r>0$ there exists $n\in \mathbb N$ and $x$ such that $|x|<r$ and $|g\_n(x)|=1.$