Practice: nice limit.

$\displaystyle \lim \limits_{x\to 0}\frac{\tan (x)}{x}=\lim \limits_{x\to 0}\left(\frac{\tan (x)-0}{x-0}\right)=\lim \limits_{x\to 0}\left(\frac{\tan(x)-\tan(0)}{x-0}\right)$ which is the definition of the derivative evaluated at $0$ . We know that: $\dfrac{\mathrm d}{\mathrm dx}\tan x=\sec^2 x.$ Therefore our limit equals: $\sec^2(0)=\dfrac{1}{\cos^2(0)}=\dfrac11=1.$

$\lim_{x \rightarrow 0} \frac{\tan x}{x} = \left( \lim_{x \rightarrow 0}\frac{\sin x}{x} \right) \left( \lim_{x \rightarrow 0} \frac{1}{\cos x} \right)$ $= 1 \cdot 1$ $= \boxed{1}$

I put no thought into this

Lim(x-->0) tgx/x=

Lim(x-->0) senx/cosx/x=

Lim(x-->0) senx/cosx.1/x=

Lim(x-->0) senx/xcosx=

Lim(x-->0) senx/x.1/cosx=

Lim(x-->0) senx/x.Lim(x-->0) 1/cosx=

1.1/cos0=

1.1/1=

1