Question

find limit as x approaches 0:
tan(x)/ (7x + sin(x))

Answer 1

Do you know l'hospital's rule?
Or do you recall that
\[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1\]

Answer 2

i recall the second one. i'd like to know how to do it with both

Answer 3

Well you have x->0
if you plug in 0 and you get an indeterminate form such as 0/0 or inf/inf then you can do l'hospital's rule
derivative of top/derivative of bottom

Answer 4

\[\lim_{x \rightarrow 0}\frac{\tan(x)}{7x+\sin(x)}=\lim_{x \rightarrow 0}=\frac{(\tan(x))'}{(7x+\sin(x))'}\]

Answer 5

OR!
\[\lim_{x \rightarrow 0}\frac{\frac{x \sin(x)}{x \cos(x)}}{7x+x \frac{\sin(x)}{x}}\]

Answer 6

I wrote tan(x) as sin(x)/cos(x)
Putting it in terms of sin(x) and cos(x) helps since we know that one limit property plus another which is \[\lim_{x \rightarrow 0}\frac{\cos(x)-1}{x}=0\]
which probably won't be used here
so we have
\[\lim_{x \rightarrow 0}\frac{\frac{x}{\cos(x)} \frac{\sin(x)}{x}}{7x+x \frac{\sin(x)}{x}} \]
Where does sin(x)/x go again as x goes to 0?

Answer 7

Do you have any questions?

Answer 8

Can you answer my question?

Answer 9

the brackets are confusing. i changed tanx to sinx/cosx, which then gives me sinx/ (cosx(7x+sinx). but i'm not sure where to go from there since it's still 0/0

Answer 10

Did you see I multiply sin(x)/cos(x) by x/x
and sin(x) by x/x

Answer 11

huh? i couldn't understand the notation. and why?

Answer 12

the notation?
the fractions?

Answer 13

yeah. it's in bracket form. i dunno. i got to 2x/ (14cosx + sin2x) and i'm stuck

Answer 14

is the code not showing up as fractions?