Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Evaluate   <img alt="\mathrm{\lim _{x \rightarrow 0}\left(\frac{(\cos (x))^{\sin (x)}-\sqrt{1-x^3}}{x^6}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%2

Evaluate   \mathrm{\lim _{x \rightarrow 0}\left(\frac{(\cos (x))^{\sin (x)}-\sqrt{1-x^3}}{x^6}\right)}

Option: 1

0


Option: 2

1/4


Option: 3

2/3


Option: 4

1/2


Answers (1)

best_answer

Let,

                                \mathrm{\mathrm{L}=\lim _{x \rightarrow 0}\left(\frac{(\cos (x))^{\sin (x)}-\sqrt{1-x^3}}{x^6}\right)}

\mathrm{\begin{aligned} & \Longrightarrow \mathrm{L}=\lim _{x \rightarrow 0} \frac{e^{\sin (x) \times \ln (\cos (x))}-e^{\frac{\ln \left(1-x^3\right)}{2}}}{x^6} \\\\ & \Longrightarrow \mathrm{L}=\lim _{x \rightarrow 0} e^{\frac{\ln \left(1-x^3\right)}{2}} \times\left(\frac{e^{\sin (x) \times \ln (\cos (x))-\frac{\ln \left(1-x^3\right)}{2}}-1}{x^6}\right) \\ & \Longrightarrow \mathrm{L}=\lim _{x \rightarrow 0} \frac{e^{\sin (x) \times \ln (\cos (x))-\frac{\ln \left(1-z^3\right)}{2}}-1}{x^6} \end{aligned}}

\mathrm{\Longrightarrow \mathrm{L}=\lim _{x \rightarrow 0} \frac{\sin (x) \times \ln (\cos (x))-\frac{\ln \left(1-z^3\right)}{2}}{x^6}\left[\text { Using } \lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}=1\right)\right]}

Now, since the limit exists, we infer,

L.H.L = R.H.L = L

\mathrm{\Rightarrow } 2L= L.H.L + R.H.L.

\mathrm{\begin{aligned} & =\lim _{x \rightarrow 0^{-}} \frac{\sin (x) \times \ln (\cos (x))-\frac{\ln \left(1-x^3\right)}{2}}{x^6}+\lim _{x \rightarrow 0^{+}} \frac{\sin (x) \times \ln (\cos (x))-\frac{\ln \left(1-x^3\right)}{2}}{x^6} \\\\ & =\lim _{x \rightarrow 0} \frac{\sin (-x) \times \ln (\cos (-x))-\frac{\ln \left(1+x^3\right)}{2}}{x^6}+\lim _{x \rightarrow 0} \frac{\sin (x) \times \ln (\cos (x))-\frac{\ln \left(1-x^3\right)}{2}}{x^6} \\\\ & {\left[\operatorname{Using}_{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} f(0-x)\right]} \\\\ & =\lim _{x \rightarrow 0} \frac{\frac{-\ln \left(1-x^6\right)}{2}}{x^6} \\\\ & =\frac{-\left(-x^6\right)}{2 x^6}\left[\operatorname{Using} \lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=1\right] \\\\ & =\frac{1}{2} \end{aligned}}

                       \mathrm{\Longrightarrow \mathrm{L}=\frac{1}{4}} 

Posted by

Riya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 Registration LIVE: NTA JEE form at jeemain.nta.nic.in; what's new this year?
anam-khan

JEE Main 2026: Wider gap in NIT cut-offs for home and other-state quotas; outsiders need better rank
anam-khan

JEE Mains 2026: Use of calculator 'not allowed’ says NTA; cites ‘typographical error' in information bulletin

Latest Question