Question

Evaluate ${lim}_{x\to 0}\frac{7xcosx-3sinx}{4x+tanx}$

Answer 1

Given ${lim}_{x\to 0}\frac{7xcosx-3sinx}{4x+tanx}$

Dividing numerator and denominator by '$x$' we get,

$={lim}_{x\to 0}\frac{7cosx-\frac{3sinx}{x}}{4+\frac{tanx}{x}}$
$=\frac{{lim}_{x\to 0}7cosx-{lim}_{x\to 0}\frac{3sinx}{x}}{{lim}_{x\to 0}4+{lim}_{x\to 0}\frac{tanx}{x}}$
[Since, ${lim}_{x\to 0}\frac{f\left(x\right)+g\left(x\right)}{h\left(x\right)}=\frac{{lim}_{x\to 0}f\left(x\right)+{lim}_{x\to 0}g\left(x\right)}{{lim}_{x\to 0}h\left(x\right)}$]
$=\frac{7\times {lim}_{x\to 0}cosx-3{lim}_{x\to 0}\frac{sinx}{x}}{4+{lim}_{x\to 0}\frac{tanx}{x}}$
Applying the formula of ${lim}_{x\to 0}\frac{\sin x}{x}=1$ and ${lim}_{x\to 0}\frac{\tan x}{x}=1$
$=\frac{7\left(1\right)-3\left(1\right)}{4+\left(1\right)}$
$=\frac{7-3}{4+1}$
$=\frac{4}{5}$