Question

what's the difference between x and theta in these trig limits??
https://i.gyazo.com/1f2639c9faf9514f1b495e091c7cd2f4.png
Why are the answers to these limits different?? Why can't 1-cos(theta) equal sin^2x??

Answer 1

in fact you can use it.

Answer 2

at least for the first one

Answer 3

\[\lim_{t \rightarrow 0} \frac{ t ^{2} }{ 1 -\cos ^{2}t }\]

Answer 4

applying the identity: \[\lim_{t \rightarrow 0} = \frac{ t ^{2} }{ \sin^{2} t }\]

Answer 5

hint:
for second limit, we can write these steps:
\[\Large \frac{{{\theta ^2}}}{{1 - \cos \theta }} = \frac{{4{{\left( {\theta /2} \right)}^2}}}{{2{{\left\{ {\sin \left( {\theta /2} \right)} \right\}}^2}}} = \frac{{2{{\left( {\theta /2} \right)}^2}}}{{{{\left\{ {\sin \left( {\theta /2} \right)} \right\}}^2}}}\]

Answer 6

but iidk get it. 0 over o remains even so..

Answer 7

i don't have a clue how to get rid of 0/0

Answer 8

what would I do next ?

Answer 9

i know how to get the answer for the one without theta

Answer 10

but the theta has a different answer that idk why it's different

Answer 11

i see, keep in mind taht they're different functions

Answer 12

thesecond one has: 1 -cos (theta) on the denominator;
while the first one has 1-cos^2 (theta) on the denominator.

Answer 13

that's why the limits are not the same

Answer 14

Have you seen L'hospital rule ?

Answer 15

not yet

Answer 16

oooh so will i solve them just like a normal variable? i didn't see the squared in the denominator

Answer 17

yeah but remember that 0 over o is not allowed, when it comes down to figuring out limits

Answer 18

yeah i know how to solve it from there with the limits

Answer 19

nice so maybe you can teach me how to solve it without usinf l'hopitals rule