for x not equal to zero what choice of f(0) will make the functions continuous at x=0
1) f(x) = [(1+x)^1/3-1]/x
2) f(x) = [5|x|-3 tan x + sin 2x]/x

Question

for x not equal to zero what choice of f(0) will make the functions continuous at x=0 
1) f(x) = [(1+x)^1/3-1]/x 
2) f(x) = [5|x|-3 tan x + sin 2x]/x

amistre64 · Answer

take the limit as x approaches zero from the left and right ....

anonymous · Answer

would you plz show a few steps ? @amistre64

amistre64 · Answer

trying to think of a simple way to approach it ...

amistre64 · Answer

can we use taylor series?

amistre64 · Answer

squeeze thrm maybe?

anonymous · Answer

have no idea about taylor series, not taught

amistre64 · Answer

well, brute math says we table it .... approach it from the left and right and log the values

amistre64 · Answer

but that requires a lot of calculation

amistre64 · Answer

$$\sqrt[3]{1+x} = 1+\frac x3-\frac{x^2}{9}+...$$
$$\sqrt[3]{1+x}-1 = \frac x3-\frac{x^2}{9}+...$$
$$\frac{\sqrt[3]{1+x}-1}{x} = \frac 13-\frac{x}{9}+...$$
the right side zeroes out and we are left with 1/3

amistre64 · Answer

but you say you dont know that method

anonymous · Answer

the value for f(0)  for the first problem is given as 1/3  
for the second it will never be continuous !

amistre64 · Answer

yeah the second one, by graph is not a hole, but a jump;  so yeah nothing could fill it in

anonymous · Answer

ok... i understand what you are trying to explain !!
so when we have x = 0 all other terms except 1/3 will be zero

amistre64 · Answer

yes

anonymous · Answer

for the second one suppose i write statements ..what should be the statements to conclude that will never be continuous

amistre64 · Answer

well, sin(u)/u = 1 as u to 0

5|x|-3 tan x + sin 2x
------------------
              x


5|x|      3 tan x       sin 2x
---- - ------- + -------
  x             x              x


5|x|      3 sinx         sin 2x      2
---- - ------- + ------- * --
  x         cosx  x          x          2



5|x|         3          
---- - ----- + 2 
  x         cosx    


5|x|                 
---- - 3 + 2 
  x         


5|x|                 
---- - 1  as x to 0 
  x

anonymous · Answer

how did  3/cos x become 3 ? did not understand that part !!

amistre64 · Answer

so,  as x to 0,  5x/x - 1 = 4
                    -5x/x - 1 = -6

cos(0) = 1

amistre64 · Answer

since we have the limit approaching 4 and -6 at the same time, there is no point that will be able to bridge the gap

anonymous · Answer

thanks a lot ..you are a hero :)

amistre64 · Answer

youre welcome ... the first one i dont see any other way about it :/

anonymous · Answer

but i understood the first one , it was like binomial expansion .. thanks again !!

amistre64 · Answer

binomial expansion yes :)  good luck

anonymous · Answer

i am closing the question