Question

Limit please ans....Class 12th​

Answer 1

Answer :-

Answer of given limit is -2 .

Solution :-

$\displaystyle \lim_{x \to \ \frac{\pi}{2}} \{2x. \tan(x) - \frac{\pi}{ \cos(x) }$

Now firstly we have to convert the given limit in 0/0 form.

↳ Converting tanx into $\frac{\sin .x}{\cos .x}$

$\displaystyle \lim_{x \to \ \frac{ \pi}{2} } \: 2x \frac{\sin(x)}{\cos(x)} - \frac{\pi}{\cos x}$

↳ Taking cos x as common denominator .

$\displaystyle \lim_{x \to \ \frac{\pi}{2} } \frac{2x. \sin(x) - \pi }{ \cos(x) }$

↳ Now let's check whether it is 0/0 form or not.

↳ Putting x as π/2 .

$\frac{2\frac{\pi}{2} . \sin(\frac{\pi}{2})-\pi}{\cos\frac{\pi}{2}}$

↳ sin π/2 = 1

↳ cos π/2 = 0

$\frac{\pi \times 1 - \pi}{0}$

$\frac{0}{0}$

↳ Hence it is in the required form now we'll imply L - Hospital rule .

Differentiate Numerator and differentiate denominator seperately .

$\displaystyle \lim_{x \to \ \frac{\pi}{2} }\frac{2\times\:1\:\times \sin (x) + 2x \times \cos(x)-0}{-\sin(x)}$

$\displaystyle \lim_{x \to \ \frac{\pi}{2} } \frac{2 \sin(x) + 2x \cos(x) }{ - \sin(x) }$

↳ Now putting x as π/2.

$\frac{ 2 \times \sin \frac{\pi}{2} + 2 \frac{\pi}{2} \times \cos \frac{\pi}{2}}{- \sin \frac{\pi}{2}}$

$\frac{2\times1+ \pi\times 0 }{-1}$

↳ 2/-1 = -2

↳ Hence answer is -2

Answer 2

putting x=π/2

2xtanx-π/cosx=2xsinx/cosx -π/cosx={2xsinx-π}/cosx={2×π/2sinπ/2 -π}/cosπ/2

now we know,

sinπ/2=1,,,cosπ/2=0

putting this value we get={π×1-π}/0=0/0in the given question we get,the value as ,0/0which is indeterminate form..,..

so now use differenitaiation rule{2xtanx -π}/cosx={2xsinx-π}/cosx

differenitaiate (2xsinx-π)/cosx

={2×1sinx+2xcosx-0}/-sinx

now putx=π/2={2×sinπ/2-2×x×cosπ/2}/-sinπ/2=2×1/-1=-2

{easiest question}