Question
Download Solution PDFEvaluate: \(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\tan x\, -\, \sin x}{\sin^{3}x}\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
We have,
\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\tan x\, -\, \sin x}{\sin^{3}x}\)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\frac{\sin x}{\cos x}\, -\, \sin x}{\sin^{3}x}\) (∵ tan x = sin x/cos x)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\sin x(1-\cos x)}{\cos x\, \sin^{3}x}\)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{(1-\cos x)}{\cos x\, \sin^{2}x}\)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{(1-\cos x)}{\cos x\, (1-\cos^{2}x)}\) (∵ sin2 x = 1 - cos2 x)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{(1-\cos x)}{\cos x\, (1+\cos x)(1-\cos x)}\)
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{1}{\cos x\, (1+\cos x)}\)
\(\therefore \frac{1}{2}\) is the correct limit.
Last updated on Jun 20, 2025
-> The Indian Navy SSR Agniveeer Merit List has been released on the official website.
-> The Indian Navy Agniveer SSR CBT Exam was conducted from 25th to 26th May 2025.
->The application window was open from 29th March 2025 to 16th April 2025.
-> The Indian Navy SSR Agniveer Application Link is active on the official website of the Indian Navy.
.->Only unmarried males and females can apply for Indian Navy SSR Recruitment 2025.
-> The selection process includes a Computer Based Test, Written Exam, Physical Fitness Test (PFT), and Medical Examination.
->Candidates Qualified in 10+2 with Mathematics & Physics from a recognized Board with a minimum 50% marks in aggregate can apply for the vacancy.
-> With a salary of Rs. 30,000, it is a golden opportunity for defence job seekers.
-> Candidates must go through the Indian Navy SSR Agniveer Previous Year Papers and Agniveer Navy SSR mock tests to practice a variety of questions for the exam.