Algebraically, what is [ lim x-->0 ] the limit of ( x + sin(x) ) / (x) ?

Question

I completely lost. Some people have suggested L'Hospital's Rule and some other stuff that my class hasn't covered. Right now, the teacher simply wants us to solve algebraically, not using some special properties. For example, in
lim x-->0 of sin2x / x I multiplied the top and the bottom by 2, making 2sin2x/2x, can celling the sin2x/2x, and leaving 2, so the limit was = 2. Something similar has to be done with these problems, thing is, I have no clue what to do.

remember that
sinx/x = 1 and 1-cosx/x = 0

I also don't know how to algebraically solve these 2 (and by the way, this is not homework, just practice for an upcoming test)

#3:
what is [ lim x-->0 ] the limit of ( sin^2(x) ) / (x) ?

#4

what is [ lim x-->0 ] the limit of ( x^2 ) / ( 1-cos(x) ) ?

and of course, the one in the actual question,

what is [ lim x-->0 ] the limit of ( x + sin(x) ) / (x) ?

Answer 1

Important to remember the limit laws. They come in handy.

The two that you make use of here
[1] lim f(x) g(x) = lim f(x) * lim g(x)
[2] lim f(x) + g(x) = lim f(x) + lim g(x)

And you're ok with lim x->0 sin(x)/x = 1
I take it that lim x->0 x/sin(x) =1 is ok too?

For this one.
lim x->0 [x+ sin(x)]/x

Split it into two fractions
lim x->0 x/x+ sin(x)/x

Simplify
lim x->0 1+ sin(x)/x

Take each limit individually. [1] the second limit is a familar one
lim x->0 1+ lim x->0 sin(x)/x
1+ lim x->0 sin(x)/x
1+ 1 = 2

lim x->0 sin^2(x)/x

Write as... (note we're giving ourself a familiar limit again)
lim x->0 sin(x)/x * sin(x)

Using [2] we have the product of limits
lim x->0 sin(x)/x lim x->0 sin(x)
1*0 = 0

This one is kind of tricky.
lim x->0 x^2 / (1- cos(x))

Multiply by (1+cos(x))/(1+cos(x)) (this let's you use a Pythagorean identity on the bottom so you can get that familar x/sin(x)).
lim x->0 x^2 (1+cos(x)) / (1-cos^2)

From Pythagorean theorem sin^2 = 1 - cos^2
lim x->0 x^2 (1+cos(x))/sin^2(x)

Break it apart into some familiar limits
lim x->0 [1+cos(x)] * x/sin(x) * x/sin(x)

limit of products to Product of the limits
lim x->0 [1+cos(x)] * lim x->0 x/sin(x) *lim x->0 x/sin(x)

And evaluate.
2* 1 * 1 = 2
---
as for the 2 sin(2x)/2x one you had..
the sin(2x)/2x doesn't "cancel" really.
but if you take
lim x->0 2 sin(2x)/2x
and you let u = 2x..
as x ->0 u also goes to 0
lim u->0 2 sin(u)/u
which we know to be 2*1

Answer 2

It will be 1 + sin(x)/x. Since sin 0 is 0, we have 0/0, which is indeterminate, and cannot be solved without using l'Hopital's rule or the crib that you were given (specifically, that sin(x)/x approaches 1 for small x).
#3: Since sin^2 (x) = 2 sin (x) cos (x), you can use the given cribs to work this one out.
#4. Use the cosine crib.

Answer 3

The answer to the initial question is 2, but not for the reason you cite. If x is in radians, sin(x)=x for small x. So in the limit, you have 2/1 or 2.

Answer 4

for the first (( x + sin(x) ) / (x)), split up the terms in the numerator (the top of the fraction) , so you end up with
x/x +sin(x)/x
and I think you can figure it out from there

For #3, it's basically the same idea, splitting up the numerator, but with multiplication this time, so you end up with sin(x)/x * x
and I believe that would end in
1*0=0

For #4... uhh... I'm not quite sure if the rule you have posted about 1-cos(x)/x applies to the reciprocal as well, but if that is the case, split it into
x * x/(1-cos(x))
maybe

Answer 5

Its quite easy

You can write [ lim x-->0 ] the limit of ( x + sin(x) ) / (x) as

[lim x --> 0] (x/x) + [lim x --> 0] [(sin x) /x]
= [lim x --> 0] (1) + [lim x --> 0] [(sin x) /x]
= 1 + 1 = 2