Question

Calculus: Limits question! (attached)

Answer 1

\[\lim_{x \rightarrow 0}(3x^2+5cosx−5)/(2x)\]

Answer 2

well if you use lospitals rule i blieve u get zero as the limit

Answer 3

so u take derivate of top and bottom (6x-5sinx)/(2) then plug in zero for x and u will get (0/2)=0

Answer 4

so its not infinity answer is zero

Answer 5

We're not to that yet. Strictly limits right now. Is there a way to take the limit without using differentiation?

Answer 6

I'd really appreciate a good explanation. Thank you in advance.

Answer 7

are you allowed to use

\[\lim_{x\to0}\frac{\sin(x)}{x}=1\]?

Answer 8

yes, we are allowed to use the limit of (sin x)/x

Answer 9

then the solution is trivial ;)

Answer 10

What do you mean?

Answer 11

cos(x)-1/x as limit x-->0 is 0 you can yous this as well and split up the limit

Answer 12

\[\frac{5\cos(x)−5}{2x}=5\frac{\cos(x)−1}{2x}\]

\[=5\frac{\cos(x)−1}{2x}\frac{\cos(x)+1}{\cos(x)+1}=5\frac{\cos^2(x)-1}{2x(\cos(x)+1)}=5\frac{-\sin^2(x)}{2x(\cos(x)+1)}=\cdots\]

Answer 13

so your left with 3x+5/2(cos(x)-1)/x as x-->0

Answer 14

3/2x i mean but it doesnt matter cause it goes to 0
3/2x+5/2(cos(x)-1)/x as x-->0

Answer 15

Just got it. Thanks a lot! :)