Question

For this challenge, I need to find the limit.

lim as theta approaches 0 of

(sin 4 theta)/(theta + tan 7 theta)

I realize that I have to rewrite the numerator and denominator so they have the same form. However, I am not sure how to treat the constants.

Guidance would be greatly appreciated!

Thank you, friends!

Answer 1

I think the way to approach this is using l'Hopital's Rule, which states that when the limit:

\[\lim_{x \rightarrow 0} \frac{ f(x) }{ g(x) } = \lim_{x \rightarrow 0} \frac{ \frac{ d }{ dx} f(x) }{\frac{ d }{ dx } g(x) }\]

Important to distinguish that this derivation is NOT like the quotient rule, you derive the numerator and denominator seperately, and also for lim x - > +-inf

Answer 2

oh wow, sentence deleted self. you use l'hopital's rule when the limit is 0/0 or inf/inf

Answer 3

Ah! Interesting! Thank you, Euler271!

Answer 4

glad i could help :) lmk if you want me to do it or if you want to compare answer ^_^

Answer 5

For sure! Any hint on how to treat the constant? As I find the limit of sin 4 theta?

Answer 6

Lim as theta -> 0 of sin theta / theta is 1.
But I am not sure what to do with the constant. Multiply it by 1?

Answer 7

it's sin(4x) so x=0 gives sin(4 * 0) . you just leave it there.

like for f(x) = sin(2x), f(pi/2) = 0 while sin(pi/2) = 1

Answer 8

Great. Thank you so much!

Answer 9

Hi here is the step by step solution as an attachment of the given problem in the form og GIF file

Answer 10

\[\lim_{\theta \rightarrow 0} \frac{ \sin(4 \theta) }{ \theta + \tan(7 \theta) } = \frac{ 0 }{ 0 + 0 }\]

0/0 therefore l'hopital's rule holds true and we apply it:

\[ = \lim_{\theta \rightarrow 0} \frac{ 4\cos(4 \theta) }{ 1 + 7\sec^2 (7 \theta) } = \frac{ 4 }{ 1 + 7 } = \frac{ 1 }{2 }\]

Answer 11

I am not clear on how the sin 4 theta becomes zero and how tan 7 theta becomes 7.

How would I search for that? Finding the limit of a trigonometric function with a constant?