Question

how to algebraically solve as lim=> infinity
(3x+sin8x)/8x

Answer 1

do you know lhopital rule?

Answer 2

alas, no. my bc calc class hasn't gone over that yet.

Answer 3

hmm nm this does not help...

Answer 4

do you know about the squeeze theorem?

Answer 5

yep! but I have to admit I have not used it a lot. I could understand maybe

Answer 6

you can break it up...

3x/8x + (sin 8x)/8x

first is 3/8, second is 0.

yu can show this using the squeeze therom...

-1 le sin 8x le 1

-1/8x goes to 0 as does 1/8x so (sin 8x)/8x must also go to 0.

Answer 7

le is <=

Answer 8

\[-1\le sin(8x)\le1 \implies-\frac{1}{x}\le \frac{sin(8x)}{x}\le\frac{1}{x}\implies \frac{sin(8x)}{x} \rightarrow 0\]
so
\[\frac{3x+sin(8x)}{3x}=\frac{\frac{3}{x}+\frac{sin(x)}{x}}{\frac{8x}{x}}=\frac{3+\frac{sin(8x)}{x}}{8}\rightarrow\frac{3}{8}\]

Answer 9

thank you all.