How many of you are going to say "L'Hopital's Rule" without figuring out that he hasn't made it to L'Hopital's Rule yet???
OP: You should have learned the limit as x approaches zero of sinx/x
(I prefer sinx instead of sinh, since sinh has another meaning: hyperbolic sin function. But, if you like "h", knock yourself out. At least we know you're not the 2nd account of Arcadio.)
Anyway, the limit of sinx/x as x-> 0 = 1. This is generally proven by using the squeezing theorem. Consider for a moment a unit circle. Without loss of generality, draw a radius to the circle in the first quadrant. Now, from the intersection of that radius and the circle, make a triangle by drawing the 3rd side to the point (1,0). Extend the radius & draw a vertical line from the point (1,0) until it reaches the extended radius forming a larger triangle. Label the angle between the radius and the x-axis with "x"
It's pretty simple to see that the area of the sector of the circle formed by the radius is greater than the area of the smaller circle, and less than the area of the larger circle. Damn, this is hard to do without a picture.
Since the area of a circle is pi r², and this circle has a radius of 1, the area of this circle is pi. The sector of the circle is just a fractional part of the circle. If it was 30 degrees, it would be 30/360ths of the circle. But, since we never use degrees in calculus, at least not in my classroom - degrees are for Babylonians and idiots who can't think in radians. I digress again.) The portion of the circle you have is x/(2pi) So, you have that portion of pi (multiply them together) and the area of that sector is x/2. The area of the big triangle is pretty simple to figure out - (1/2)(base)(height.) The base is obviously 1, and the 8th grader sohcahtoa chant can be used to find the height, as the tangent(x) = height over bottom of the triangle (which is one.) So, the area of that triangle is tanx/2. Likewise, the area of the smaller triangle works out to sinx/2.
So, .5sinx <= x/2 <= .5tanx
Oh, I forgot to mention, I included = because what if the angle is zero, huh? If the angle is zero, the area of each triangle (and the sector) is zero.
Anyway, a little manipulation of this: multiple all three parts by 2 to get rid of the damn fractions.
sinx <= x <= tanx
Now, divide each by sinx
1 <= x/sinx <=tanx/sinx
1 <= x/sinx <= 1/cosx
Take the reciprocals of each, which will reverse the order of the inequalities (1/3 is less than 1/2, but 3 is greater than 2)
1 >= sinx/x >= cosx
So, the function f(x)=sinx/x will lie between 1 and the function g(x)=cosx
As x approaches zero, the limit of 1 is 1, the limit of cosx is also 1.
What value of sinx/x works for this:
1 >= sinx/x >= 1??? The answer is: 1
So, now that I proved that the limit of sinx/x as x->0 is one, let's use it.
What's the limit as x ->0 of sin7x/sin6x? Damn, that doesn't look like what I just proved.
Well, multiply the top and bottom of that fraction by 1/x. No problem since x never actually equals zero.
lim [sin7x/x] / [sin6x/x]
heyyyyy, it's looking closer.
Multiple the fraction in the numerator by 7/7, and the fraction in the denominator by 6/6
[7sin(7x)/(7x)]/[6sin(6x)/(6x)]
Now, if you know the limit as x->a of f(x) and the limit as x->a of g(x), then the limit of (f(x))(g(x)) is just the product of the limits, limit of the sums is the sum of the limits. etc. (as long as you're not dividing by zero.)
So, the sin(7x)/(7x) part approaches 1, the sin(6x)/(6x) part also approaches 1. So, you have 7 times 1 divided by (6 times 1) = 7/6
Sidenote: with L'hopital's rule, this problem would take about, ohhhh, 2 seconds.
Now, for your problems, you need to manipulate the function inside the limit so that part of it matches the pattern (sin(some angle))/(some angle) That part of the function approaches 1 as x approaches zero.
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