Question

Limits

The value of f(0) so that the f(x) is continuous at x=0.

Answer 1

and f(x)=?

Answer 2

\[\Large f(x)=\frac{(27-2x)^{1/3}-3}{9-3(243+5x)^{1/5}}\]

Answer 3

L hospital B)

Answer 4

ok

Answer 5

trying

Answer 6

\[\Large f'(x)=\frac{\frac{1}{3}(27-2x)^\frac{-2}{3}\times -2}{-3(\frac{1}{5}(243+5x)^\frac{-4}{5}\times 5} \]

Answer 7

try subbing x=0 now

Answer 8

i am getting 37.44

Answer 9

and ans. is 2

Answer 10

im getting 2 only,recheck ur calculation

Answer 11

ok

Answer 12

\[\Huge \frac{\frac{-27}{2}}{\frac{1}{-27 \times 5} \times 5}\]

Answer 13

sorry its 2/27 in numerator,typo*