Question

2 questions in calculus:

Answer 1

Why not one at a time?

Answer 2

prove:
1) \[\lim_{x \rightarrow 0+} \frac{x}{x-\left[ sinx \right]} = 1\]
and
\[\lim_{x \rightarrow 0-} \frac{x}{x-\left[ sinx \right]} = 0\]

2) Prove:
\[\lim_{x \rightarrow \frac{ \Pi }{2}}\frac{1}{cosx} \neq \infty\]

Answer 3

my bad abb0t, I believe these are 2 easy ones... but I'll ask 1 at a time from now on

Answer 4

Multiply the first one by the conjugate

Answer 5

Also, \(\frac{ 1 }{ \cos(x) } = \sec(x)\)

Answer 6

we're supposed to prove using the epsilon/delta definition of limit

Answer 7

and I can't seem to finish it... got question 1 to this:
\[|\frac{sinx}{x-sinx}|\]
and now I need to define Delta somehow

Answer 8

For every \(\epsilon >0\) there exists some \(\delta >0\) such that \(|f(x)-L|< \epsilon \) whenever \(0<|x-a|< \delta\)

Answer 9

L in the first case is 1

Answer 10

and in question 2 I can't find how to define M so \[\frac{1}{cosx}\le N\]

Answer 11

i know how to define but I have a problem with the steps you take to figure out how to pick the specific delta/epsilon

Answer 12

\(\left| \frac{ x }{ x-\sin(x) } - 1 \right| < \epsilon\) whenever \(0<|x-0|< \delta \)

do the same with zero now. Use algebra to solve for epsilon basically.

Answer 13

since epsilon is just some arbitrary number.

Answer 14

I did this and got to
\[|\frac{sinx}{x-sinx}|\]
when I know that |x|<delta
and now i'm stuck on how to define delta so this expression will be smaller than epsilon

Answer 15

I usually get stuck on changing the expresion using algebra

Answer 16

same with the other question, I know what I'm supposed to do but I'm stuck on how to 'play" with equation