What is the limit (calculus)?

Question

these are in my calculus textbook and i can't figure them out

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

and


lim (sin(x+pi)) / x
x->0

hey Pascal, thanks, but we haven't learned L'Hopital's rule yet!

Answer 1

If you are permitted to use L'hopital's rule, these are simple. Substituting, we find that both of these limits yield the indeterminate form 0/0, so we have:

[x→0]lim sin (-2x)/x = [x→0]lim -2 cos (-2x)/1 = -2

[x→0]lim sin (x+π)/x = [x→0] cos (x+π)/1 = -1

If you have to evaluate these without L'hopital (which I suspect, if you can't figure them out), then this is slightly harder, but still possible. Taking the first one:

[x→0]lim sin (-2x)/x
Using sin (-x)=-sin x:
[x→0]lim -sin (2x)/x
Applying the double angle formula:
[x→0]lim -2sin x cos x/x
Rearranging:
[x→0]lim -2 cos x sin x/x
We know that [x→0]lim sin x/x = 1 (this may be proven geometrically, without L'hopital, and in fact is an integral part of the proof that d(sin x)/dx = cos x). Thus, considering sin x/x as a single unit, this limit assumes the determinate form -2 * 1 * 1, so:
[x→0]lim sin (-2x)/x = -2

For the second:
[x→0]lim sin (x+π)/x
Using the angle addition formula:
[x→0]lim (sin x cos π + cos x sin π)/x
Simplifying:
[x→0]lim -sin x/x
And because [x→0]lim sin x/x = 1:
[x→0]lim sin (x+π)/x = -1

Answer 2

Use L'Hospital's Rule on both of them. If we let x go to zero in both the numerator and denominator of both problems, the result is 0/0, an undefined operation in mathematics. L'Hospital's Rule says that when this is the case, then we take the derivative of both the numerator and denominator, and then substitute the limit in them to get a result. Sometimes we must take more than one derivative before an undefined condition is eliminated.

In the first problem, sin(-2*0) = sin(0) = 0, and x = 0, so we have the undefined condition, 0/0, present here. Using L'Hospital's Rule, we first take the derivative of top and bottom. In the numerator, we use the Chain Rule and find du/dy*dy/dx. So du/dx is:

d(sin(-2x)) *d(-2x) = cos(-2x)*(-2)= -2cos(-2x).

Subsituting the limit, 0, in for x we get:

-2cos[-2(x)] = -2cos[-2(0)] = -2cos[0] = -2[1] = -2.

Taking the derivative of the denominator, d(x)/dx = 1. Since this is a constant, when x equals 0, the denominator still equals 1.

Now all we do is divide the numerator by the denominator to get:

-2/1 = -2, which is the answer to problem 1.

For the second problem, we have a similar situation. Letting x go to zero in the numerator and denominator results in the undefined condition, 0/0, so this calls for the use of L'Hospital's Rule again.

Taking the derivative of sin(x+pi) by again using the Chain Rule, we get:

cos(x+pi)*(1+0) = 1*cos(x+pi) = cos(x+pi).

Letting x assume the value 0, we get:

cos(0+pi) = cos(pi) = -1.

The derivative of the bottom is again 1.

Divding the numerator by the denominator yields -1/1 = -1, which is the answer to problem two.

Answer 3

it somewhat is an outline of the belief of a decrease in arithmetic. For particular makes use of of a decrease, see decrease of a chain and decrease of a function. In arithmetic, the thought of a "decrease" is used to describe the value that a function or sequence "techniques" because of the fact the enter or index techniques some value. Limits are needed to calculus (and mathematical diagnosis many times) and are used to outline continuity, derivatives, and integrals. the thought of a decrease of a chain is extra generalized to the thought of a decrease of a topological internet, and is heavily regarding decrease and direct decrease in classification thought. In formula, decrease is often abbreviated as lim as in lim(an) = a, and the certainty of coming near a decrease is represented with the aid of the the final option arrow (?) as in an ? a.

Answer 4

Thats odd they are in my Algbra book Oh I don't know

Answer 5

Sorry but I think you need to do your own homework!