Question

find the limit of (3x-sin(kx))/x as x approaches zero. k is not equal to zero.
What I did was multiply the top and bottm by k. so that way sin(kx)/kx would equal one. that left me with 3x-1*k and that got me 3x-k. but my book says the answer is 3-k. can someone explain this to me

Answer 1

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{3x-\sin(kx)}{x}}\)

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{3x}{x}-\lim_{x \rightarrow ~0}\frac{\sin(kx)}{x}}\)

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}3~-\lim_{x \rightarrow ~0}\frac{k~\sin(kx)}{kx}}\)

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}3~-k\lim_{x \rightarrow ~0}\frac{\sin(kx)}{kx}}\)

Answer 2

from here apply your theorem that \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{\sin(x)}{x}=1}\)

Answer 3

So at your last step, you will get that,

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}3-k\times 1}\)

\(\large\color{blue}{\displaystyle 3-k}\)

Answer 4

oh okay and on the part where its k *sinkx the limit of k is k right? because it is constant? @SolomonZelman

Answer 5

well, I took the constant out, but yes, if you are taking the limit of the function as X approaches zero, and you have only the k inside, then it remains k.

Answer 6

In fact, \(\large\color{blue}{\displaystyle \lim_{x \rightarrow ~a}k=k}\) for any (real) number a.

Answer 7

because you are not plugging anything for x.

Answer 8

is everything clear ?

Answer 9

oooo thank you! and one last question! why were you able to separate the 3x from sinkx?

Answer 10

because the rule is:

\(\large\color{blue}{\displaystyle \lim_{x \rightarrow ~a}~\left({\huge\color{white}{|}} f(x)-g(x) {\huge\color{white}{|}}\right)=\lim_{x \rightarrow ~a}f(x)-\lim_{x \rightarrow ~a}g(x)}\)

Answer 11

same thing about multiplication, division and a sum of limits.

Answer 12

ah yes thank you its because I was only thinking of division lol

Answer 13

\(\large\color{slate}{\displaystyle \lim_{x \rightarrow ~a}\left[ f(x)^{\color{white}{|}}\right]^b= \left[\lim_{x \rightarrow ~a}f(x) \right]^{b}}\)

Answer 14

limits, (unlike integrals) are very manipulable in this sense

Answer 15

yes, separating the limits is extremely important in solving limits