Ask your own question, for FREE!
Calculus1 18 Online
OpenStudy (wesdg1978):

lim x-->0 sin5x/x How do I find the limit using: lim x=0 sinx/x=1

hartnn (hartnn):

you need same angle to tend to 0, which is with sine function, and the same denominator, so, when x-->0, 5x also -->0 for denominator, you can do this adjustment, \(\large 5 \dfrac{\sin 5x}{5x}\)

hartnn (hartnn):

now you have, \(\large 5\lim \limits_{5x \rightarrow 0 } \dfrac{\sin 5x}{5x}\) which is of the standard form. can you solve now ?

OpenStudy (wesdg1978):

To me, that now looks like this: 5 lim 5x-->0 sin 1 So is it 1/5?

hartnn (hartnn):

no.... |dw:1371411201813:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!