Question

Solve:

Limx-->0+ sinx/(x^2)
Limx-->0- sinx/(x^2)

To prove Lim x-->0 sinx/(x^2) DNE without using L'Hopital's Rule

Answer 1

How would I do this:

1. Without using squeeze theorem

2. With squeeze theorem?

Answer 2

without:

plug 0+ (0.0000001) and 0- (0.0000001) in your calculator to prove that lim 0+ DNE 0- therefore limit DNE.

with:

\[-1 \le sinx \le 1\]\[-\frac{ 1 }{ x^2 }\le\frac{ sinx }{ x^2 }\le \frac{ 1 }{ x^2 }\]\[lim_{x->0}(-\frac{ 1 }{ x^2 })\le\frac{ sinx }{ x^2 }\le lim_{x->0}(\frac{ 1 }{ x^2 })\]
\[- \infty \le \lim_{x->0} \frac{sinx}{x^2} \le \infty\]

since left and right aren't the same. we can not deduce the limit of sinx/x^2 by squeeze theorem since it DNE

Answer 3

0- = (-0.0000001)

Answer 4

the 3rd line of squeeze theorem should be lim of sinx/x^2 as well.

we took the limit of all 3 terms at the same time for this to be true.


feel free to ask questions ^_^

Answer 5

I get it! Thank you! How would you solve this without squeeze theorem?

I'm actually trying to find the vertical asymptote using limits!

Answer 6

by graphing. should look like this at zero:



or plug in 0- and 0+ on your calculator.

0- means 0 coming form the left. the closest number to zero from the left is -0.0000000000000.........1

and 0+ is coming from the right of zero. so you can plug in +0.0000000....1

with you calculator, not that many 0s are necessary to estimate if you get the same result or not