# Show that if lim_(x->a) f(x)=L, then lim_(x->a) cos(f(x))=cos(L).

Show that if $\underset{x\to a}{lim}f\left(x\right)=L$, then $\underset{x\to a}{lim}cos\left(f\left(x\right)\right)=cos\left(L\right)$.
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Mackenzie Lutz
$\mathrm{cos}\left(A\right)-\mathrm{cos}\left(B\right)=-2\mathrm{sin}\left(\frac{A+B}{2}\right)\mathrm{sin}\left(\frac{A-B}{2}\right)$
Apply this with $A=f\left(x\right)$ and $B=L$L. Then
$|\mathrm{cos}\left(f\left(x\right)\right)-\mathrm{cos}\left(L\right)|=2|\mathrm{sin}\left(\frac{f\left(x\right)+L}{2}\right)\mathrm{sin}\left(\frac{f\left(x\right)-L}{2}\right)|.$
Now, recall that $|\mathrm{sin}\left(t\right)|\le |t|$, so,
$|\mathrm{cos}\left(f\left(x\right)\right)-\mathrm{cos}\left(L\right)|\le \frac{1}{2}|f\left(x\right)+L||f\left(x\right)-L|.$
And from here, it is easy to conclude a $\delta -\epsilon$ type proof.
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Ivan Buckley
For the sake of completeness,
$\mathrm{cos}\left(a+b\right)=\mathrm{cos}\left(a\right)\mathrm{cos}\left(b\right)-\mathrm{sin}\left(a\right)\mathrm{sin}\left(b\right)$
$\mathrm{cos}\left(a-b\right)=\mathrm{cos}\left(a\right)\mathrm{cos}\left(b\right)+\mathrm{sin}\left(a\right)\mathrm{sin}\left(b\right)$
by substracting
$\mathrm{cos}\left(a+b\right)-\mathrm{cos}\left(a-b\right)=-2\mathrm{sin}\left(a\right)\mathrm{sin}\left(b\right).$
Now, calling $A=a+b$ and $B=a-b$, gives $a=\frac{A+B}{2}$ and $b=\frac{A-B}{2}$.