Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The value of f(0), so that the function   <img alt="\mathrm{f(x)=\frac{(27-2 x)^{1 / 3}-3}{9-3(243+5 x)^{1 / 5}}(x \neq 0)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

The value of f(0), so that the function   \mathrm{f(x)=\frac{(27-2 x)^{1 / 3}-3}{9-3(243+5 x)^{1 / 5}}(x \neq 0)}  is continuous, is given by

Option: 1

\frac{2}{3}


Option: 2

6


Option: 3

2


Option: 4

4


Answers (1)

best_answer

Since f(x) is continuous at x=0, therefore
\mathrm{ \left.f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{1 / 3}-3}{9-3(243+5 x)^{1 / 5}} \quad \text { (Form } \frac{0}{0}\right) }

\mathrm{ =\lim _{x \rightarrow 0} \frac{\frac{1}{3}(27-2 x)^{-\frac{2}{3}}(-2)}{\frac{3}{5}(243+5 x)^{-\frac{4}{5}}(5)}=\left(-\frac{2}{3}\right)\left(-\frac{1}{3}\right) \frac{3^4}{3^2}=2 . }

Posted by

Ritika Harsh

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 correction window opens tomorrow; NTA allows exam city change
anam-khan

Faridabad: 17-year-old girl preparing for JEE Main 2026 dies by suicide
anam-khan

NTA JEE Mains 2026 application closes at 9 am today; correction facility from December 1

Latest Question