Question

again. in limit!

find a non zero value for the constant k that makes
f(x) = tan kx/x , x < 0
3x + 2k^2 , x is bigger and equal to 0

continuous at x =0.

Answer 1

tried it ? any ideas how to start ?

Answer 2

i've tried but no solution.
what i did is
i substitute the x to 0 but i got 0 for each function

and yet, i try to make it like this
\[\frac{ \tan kx }{ x } = 3x + 2kx ^{2}\]

and no solution too

Answer 3

since the function is given continuous at x = 0

\(\Large \lim \limits_{x\to 0^-}f(x)=\lim \limits_{x\to 0^+}f(x)\)

Answer 4

x approaches 0- means x is very near to 0, but LESS than 0
so you would use the function
tan kx/x when x approaches 0-
makes sense ?

Answer 5

because f(x) is tan kx/x with x<0

Answer 6

\(\Large \lim \limits_{x\to 0^-}\dfrac{ \tan kx }{ x }=\lim \limits_{x\to 0^+} 3x + 2kx ^{2}\)


now solve those 2 limits individually

Answer 7

the right side limit is very simple, you just plug in x=0

Answer 8

left side limit,
use
tan = sin/cos

\(\Large \lim \limits_{x\to 0} \dfrac{\sin x}{x} =1\)

Answer 9

***
\(\Large \lim \limits_{x\to 0^-}\dfrac{ \tan kx }{ x }=\lim \limits_{x\to 0^+} 3x + 2k ^{2}\)

Answer 10

ohhh almost forgot about that. i got the clue. let me try first. =)

Answer 11

but... i got 0. hmmm?

Answer 12

right side = 2k^2
any doubts in that ?

Answer 13

show your try for left side limit.

Answer 14

but it is actually 2kx^2

Answer 15

you mis-typed the question ?

then yes, right side =0

Answer 16

how about left side limit ?

Answer 17

got 0 too. because tan 0=0

Answer 18

tan 0/0 = 0/0 form


use
tan = sin/cos

Answer 19

\[\frac{ \frac{ \sin kx }{ \cos kx } }{ x }\]

thats true?

Answer 20

yesss

which is just

sin kx/ (x cos kx)

Answer 21

the answer still zero?

sorry , i'm a bit slow in limit.

Answer 22

yes, k is 0
but the left hand limit is NOT 0

Answer 23

LHL = 1?