Question

Evaluate the limit:

$\underset{x\to 0}{lim}\frac{1-cos2x+tan{}^{2}x}{xsinx}$

Answer 1

We have,

$\underset{x\to 0}{lim}\frac{1-cos2x+tan{}^{2}x}{xsinx}$

It becomes 0/0 form.

$=\underset{x\to 0}{lim}\frac{2si{n}^{2}x+tan{}^{2}x}{xsinx}[\because 1-cos2x=2si{n}^{2}x]$

On dividing numerator and denominator by $x^2$,

we get

$=\underset{x\to 0}{lim}\frac{\frac{2si{n}^{2}x}{x^2}+\frac{tan{}^{2}x}{x^2}}{\frac{xsinx}{x^2}}$

$=\underset{x\to 0}{lim}\frac{2{\left(\frac{sinx}{x}\right)}^2+{\left(\frac{tanx}{x}\right)}^2}{\frac{sinx}{x}}$

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

$=\frac{\left\{2\right(1{)}^2+\left(1{)}^2\right\}}{1}$

$=3$

Therefore,

$\underset{x\to 0}{lim}\frac{1-cos2x+tan{}^{2}x}{xsinx}=3$