Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{7xcosx-3sinx}{4x+tanx}$

Answer 1

We have,

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

It becomes 0/0 form.

Dividing numerator and denominator by $x$ , we get

$=\underset{x\to 0}{lim}\frac{7cosx-3\left(\frac{sinx}{x}\right)}{4+\left(\frac{tanx}{x}\right)}$

$=\frac{7\underset{x\to 0}{lim}\left(cosx\right)-3\underset{x\to 0}{lim}\left(\frac{sinx}{x}\right)}{4+\underset{x\to 0}{lim}\left(\frac{tanx}{x}\right)}$

$[\because \underset{x\to 0}{lim}\frac{sinx}{x}=1,\underset{x\to 0}{lim}\frac{tanx}{x}=1]$

$=\frac{7\cdot 1-3\cdot 1}{4+1}$

$=\frac{4}{5}$

Therefore,

$\underset{x\to 0}{lim}\frac{7xcosx-3sinx}{4x+tanx}=\frac{4}{5}$