Introduction to limits

Transcription

1 Introduction to limits 9 Secants and Tangents Let f be a function whose domain/codomain are intervals D and C. 9. Secant slope For r, s D, the secant line is the one through the graph points (r,f(r)), and (s,f(s)). The secant slope is: f(s) f(r) slope = s r secant line equation: y = ( f(s) f(r) s r ) (x r) + f(r) Physical interpretation: Suppose a function d(t) is measuring the position of a moving point (on an axis) for input time. Then: d(s) d(r) s r = change in position between the two times t = s and t = r, divided by the change in time, has the physical meaning of average speed.

2 9.2 The example f(x) = x 2 Take f(x) = x 2. The slope of the secant line for the two graph points P = (r,r 2 ) and Q = (v,v 2 ) is: slope = v2 r 2 v r = v + r A picture of some secant slopes for P = (0.5,(0.5) 2 ) is:

3 Tangent slope at specific point P = (0.5,0.25): We fix r = 0.5 and consider let v tend to 0.5. The secant slope of the line from P = (0.5,0.25) to Q = (v,v 2 ) is v As v tends to 0.5, the secant slope will tend to =. The tangent slope at point P is. Tangent slope at general point P = (b,b 2 ): More generally, we can compute the slope of the tangent line to the graph point Q = (b,b 2 ) the same way: As we let we get secant slope from P = (b,b 2 ) to Q = (v,v 2 ) is: v b, (v +b) (b+b). v 2 b 2 v b = v + b So, the tangent slope to the graph point P = (b,b 2 ) is 2b.

4 9.3 The example g(x) = x 3 Take g(x) = x 3. The slope of the secant line for the two graph points P = (b,b 3 ) and Q = (v,v 3 ) is: slope = v3 b 3 v b = (v b)(v2 +vb+b 2 ) (v b) = (v 2 +vb+b 2 ) As we let we get v b, secant slope b 2 +bb+b 2 = 3b 2. So, the slope of the tangent line to the graph at the graph point P = (b,b 3 ) is 3b 2.

5 0 Summary of tangent slope as a limit Let f be a function whose domain/codomain are intervals D and C. Secant line: For b, v D, the secant line is the one through the graph points P = (b,f(b)), and Q = (v, f(v)). The secant slope is: secant line PQ slope = f(v) f(b) v b Slope of tangent line at P = (b,f(b)): We wish to take a limit as v b, to find the slope of the tangent line at P: lim v b f(v) f(b) v b An algebraic example: We take the algebraic rule f(x) = x on the interval (0, ). For P = (b, b ) and Q = (s, s ). v b v b vb = b 2 secant line PQ slope = lim v b = b vb v vb v b = b v vb(v b) = vb

6 The slope of the tangent line to the graph of f(x) = x at the point P = (b, b ) is b 2. Alternate but equivalent way: For f a function, we let h be the difference v b between the inputs s and b; so v = b+h. Then: f(b+h) f(b) secant line PQ slope = h f(b+h) f(b) slope of tangent line at point P = lim h 0 h

7 Example: We take the rule f(x) = x = x 2. For b > 0, find the tangent slope to the graph of f at the point P = (b,b2). We take Q = ((b+h),(b+h) 2), and calculate the secant slope: ( secant slope = (b+h) 2 b 2 = (b+h) 2 b ) 2 (b+h) 2 + b 2 ( ) h h (b+h) 2 + b 2 = (b+h) b h = ( (b+h) 2 + b 2 ( (b+h) 2 + b 2 As h 0, we have (b+h) b, and ( ), so: (b+h) 2 + b2 2b 2 ) ) tangent slope to graph at P = 2b 2.

8 Limits Definition of when a function has a limit. Suppose D is an interval, and f is a function whose domain is D with the possible exception of an interior point b. For example, for the function y = x 2, and P = (b,b 2 ), the secant slope of the line P and Q = (x,x 2 ) is m P (x) = x2 b 2 x b Inthisalgebraicexpression(form P )bewemustexcludeb division by zero is not allowed.

9 . We say a function f has a limit L as x b if: Examples. st Intuition formulation of limit: We can assure the output values f(x) are close to L by taking the input x to be close to but not equal to b. lim x b x 3 = b 3. Our intuition says if we take x near to b, then x 3 should be near to b 3. For the function y = x 2, since the secant slope of P = (b,b 2 ) and Q = (x,x 2 ) is x2 b 2 x b, and m P (x) = x2 b 2 x b = x + b, if we now take the limit of the secant slope as x b we get: lim x b x 2 b 2 x b = lim x b (x + b) = 2b.

10 .2 2nd More quantitative formulation of limit: If we take x near to (but not equal to) b; so 0 < x b is small, then f(x) will be near to L, that is f(x) L is small. Example. We use this 2nd definition of limit to show lim x b x = b. The limit value here is L = b. We have x b = x b x + b x + b = x b x + b Therefore, if we make x b small, the quantity x b = x b x + b will be small too.

11 .3 Formulation of limit in a quantitative manner: 3rd Quantitative formulation of limit: Given a challenge to make the quantity f(x) L small, say smaller than some tolerance T, we can find a tolerance-reply positive number R with the property that 0 < x b < R implies = f(x) L < T. Examples. We use the quantitative definition of limit to show: lim x b x = b. We will asssume b 0. We calculated above that x b = x b x + b. Since b ( x+ b), wehave x b x b b. Givenachallengetomake x b < T, we see we can do so by taking x b < R = T b.

12 lim x b x 2 = b 2. We will asssume b > 0. We algebraically manipulate x 2 b 2 to get x 2 b 2 = (x b) (x + b) = x b x + b If we are presented with a tolerance T > 0 and challenged to make x 2 b 2 < T, we can do so by insuring two things (since b > 0): (i) Make x b less than T 2b, and (ii) make x + b less than 2b. The first is the requirement x b < T 2b. The second means (since b > 0) that 2b < x + b < 2b so 2b b < x b < 2b b = b Now, 3b < (x b) < b will be true when x b < b. (Note. x b < b implies 3b < (x b) < b but is not equivalent to it.) We can make both true by taking x b < T 2b and x b < b. This means take x b less than both T 2b and b. So our reply R to the challenge T (to make x2 b 2 < T) is to take x b < minimum of R = minimum of T 2b and b T and b. 2b