Q102 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
$\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi(x+2)\right)}{x^2}=$
- A$\frac{\pi}{2}$
- B$\frac{\pi^2}{4}$
- Correct answerC$\frac{\pi^2}{2}$
- D$\frac{\pi}{4}$
Solution
We have,
$\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi(x+2)\right.}{x^2}$
$\lim _{x \rightarrow 0} \frac{1-\cos \left(2 \pi+\pi x+x^2\right)}{x^2}$
$\lim _{x \rightarrow 0} \frac{1-\cos \left(\pi x+x^2\right)}{x^2} \quad[\because \cos (2 \pi+\theta)=\cos \theta]$
Applying 'L' Hospital rule
$\lim _{x \rightarrow 0} \frac{\sin \left(\pi x+x^2\right)(2 x+\pi)}{2 x}$
$=\lim _{x \rightarrow 0} \sin \left(\pi x+x^2\right)+\lim _{x \rightarrow 0} \frac{\pi \sin \left(\pi x+x^2\right)}{2 x}$
$=0+\frac{\pi}{2} \lim _{x \rightarrow 0} \frac{\sin \left(\pi x+x^2\right)}{x}$
Again apply 'L' Hospital rule
$=\frac{\pi}{2} \lim _{x \rightarrow 0} \frac{\cos \left(\pi x+x^2\right)(2 x+\pi)}{1}=\frac{\pi}{2} \times \pi=\frac{\pi^2}{2}$
Hence, the correct answer is option: 'C'.