Click to See Complete Forum and Search --> : Need a little help with a limit/trig. problem

11-27-2002, 01:49 PM
We just started to do this type of problem in my Calculus class and I am hoping that someone can help me with these two problems:

Recall that lim(as x->0) (sinx)/(x) = 1 Using this fact, compute

1. lim(x->0) (sin7x)/(11x)

2. lim(x->0) (tan2x)/(x)

11-27-2002, 09:24 PM
1.) If you are in Calculus I, then you will probably have to apply the "Squeeze Theorem" to find the limit of sin(7x)/11x as x approaches zero. If you're in Calculus II, then you can simply use L'Hopital's rule to find the limit.

Calculus I:
(There are different ways of showing how this can can be accomplished, here's one way)

Like your problem says lim(as x->0) sin(x)/x = 1, notice that there is an 'x' in sin(x) and 'x' in the denominator. mmkay? My point is 'x' in both have the same coeffecients (Which is 1). Hmmmm?

lim -------- =
x->0 11x

Let's factor out the 11 from the denominator and rewrite our equation like this:

1 / sin(7x) \
lim -- | ------- | =
x->0 11 \ x /

Look inside the parenthesis. You see sin(7x)/x. Let's make the 'x's look the same. Meaning, let's give the denominator the same coeffecient as what's inside the sine function "sin(7x)".

7 / sin(7x) \
lim -- | ------- | =
x->0 11 \ 7x /

What I really did was multiply by one. 7/7 = 1 Right? Okay... Well! Look at that, both 'x's have the same coeffecients of 7! That means we can say sin(7x)/7x = 1. It's just like saying sin(x)/x = 1. As long as the coeffecients of the 'x' are the same for both the sine function and the denominator. So

lim ------ = 1, then we can say
x->0 x

lim ------- = 1. Which means
x->0 7x

7 / sin(7x) \
lim -- | ------- | =
x->0 11 \ 7x /

(Use the product property of limits)
7 sin(7x)
lim -- * lim ------- =
x->0 11 x->0 7x

7 7
lim -- * 1 = --
x->0 11 11

I know my explaination is long, but the problem really isn't very long when worked out.

Calculus II:
(L'Hopital's Rule. Makes life much simpler)

lim ------- =
x->0 11x

lim -------------- =
x->0 d[11x]

7cos(7x) 7
lim -------- = --
x->0 11 11

2.)Pretty much the same as above. For Calculus I, use the trig. identity: tan(x) = sin(x)/cos(x). For your problem it will be written as: tan(2x) = sin(2x)/cos(2x). Do the same thing as above by factoring out the cosine from the denominator. Then find a coeffecient for the 'x' under the sine to match the coeffecient of the 'x' in the sine (sin(2x)). The answer will be 2.

Good Luck. I hope I helped and didn't scare you away :D

And let me know if you figure out number 2. I'd like to know how you do.