Solving limits with trig. functions

[mybrainhurts1]

Can anyone help me with these two limit problems...
solve:

lim h->o (h^2-h+sinh)/2h

and

lim y->0 (sin3cot5y)/ycot4y

I know lim x->0 sinx/x=1, but I can't figure out how to get these two into that format.

 

[Mark44]

Can anyone help me with these two limit problems...
solve:

lim h->o (h^2-h+sinh)/2h

Split it up into the sum of three separate limits.

and

lim y->0 (sin3cot5y)/ycot4y

Typo here? Should this be lim y->0 (sin3ycot5y)/ycot4y?
If so, split it up into the product of three separate limits, after working with cot5y and cot4y.

I know lim x->0 sinx/x=1, but I can't figure out how to get these two into that format.

 

[mybrainhurts1]

Yes, should be lim y->0 (sin3ycot5y)/ycot4y

So for the first problem--
(h^2/1)-(h/2)+sinh/h= (0^2/1)-(0/2)+1=1...?

 

[Mark44]

No.
First, you omitted the fact that a limit is involved.
Second, you have multiple algebra errors.

 

[mybrainhurts1]

You're very helpful.

 

[Mark44]

Are you able to find the errors in the first problem?

 

[mybrainhurts1]

limx->o h^2/2h - limx->0 h/2h + limx->0 sinh/2h, but where do I go from here.

 

[Mark44]

\lim_{h \to 0} \frac{h^2}{2h} = (1/2) \lim_{h \to 0} h * \frac{h}{h}=(1/2) \lim_{h \to 0} h * \lim_{h \to 0}\frac{h}{h}

It's valid to move constants in or out of the limit, and it's valid to split up limits into sums or products of limits, as long as the individual limits exist.

Can you continue from here for this limit and the other two?