Question

Let f(x) = (sin(4x))/x and g(x) = 3x2 + 2.

Find the limit of f(x) + g(x) as x approaches 0.

Answer 1

the limit in \(g\) is easy right? replace \(x\) by \(0\)

Answer 2

yeah, its the limit of f i'm confused on

Answer 3

so the real question is, what is
\[\lim_{x\to 0}\frac{\sin(4x)}{x}\]

Answer 4

don't you replace it with cosine?

Answer 5

nah

Answer 6

gimmick is to multiply top and bottom by \(4\)

Answer 7

\[4\lim_{x\to 0}\frac{\sin(4x)}{4x}=4\times 1\]

Answer 8

this assumes you know \(\lim_{x\to 0}\frac{\sin(x)}{x}=1\) which you do

Answer 9

if you want to waste time you can use l'hopital but it is not necessary here
for that matter, you can recognize
\[\lim_{x\to 0}\frac{\sin(4x)}{x}\] as the derivative of \(\sin(4x)\) evaluated at 0, i.e. \( 4\cos(0)\)

Answer 10

Or you could directly use L'Hopital and take \[\lim_{x->0} 4\cos(4x)/1 = 4\]

Answer 11

or you could do what @whpalmer4 said

Answer 12

@whpalmer4 \to

Answer 13

so essentially just 4 and then 2 from g?

Answer 14

\(x\to 0\)

Answer 15

right

Answer 16

cool, thanks!

Answer 17

It's all what you remember better...beginners tend to remember the derivative of sin before they remember the limit of sin x/x, people with some EE background remember the sinc function

Answer 18

true, but often these problems come pre l'hopital, and right after the limit of sine, usually used to prove the derivative to begin with

Answer 19

hmm, as I recall every calc book I ever used did L'Hopital very early on, before tackling trig functions. but no matter, they're in storage, and she isn't using them :-)

Answer 20

Mathematica calculated 6 as the limit.
Refer to the attachment.

Answer 21

yeah, 6 is what we got, too...we decided the limit of the right hand term was 2 right off the bat and concentrated on the left hand term