1.5 Inverse Trigonometric Functions Remember that only one-to-one functions have inverses. So, in order to find the inverse functions for sine, cosine, and tangent, we must restrict their domains to intervals where they are one-to-one. To find the inverse sine function, we restrict the domain of sine to [−π=2; π=2]. sin−1 x = y $ sin y = x In other words, sin−1 x or arcsin x is the ANGLE in the interval [−π=2; π=2] whose sine is x. p −1 3 1 −1 sin 2 arcsin(− 2 ) sin 2 In order to have an inverse for cosine, we restrict the domain of cosine to the interval [0; π]. cos−1 x = y $ cos y = x In other words, cos−1 x or arccos x is the ANGLE in the interval [0; π] whose cosine is x. p −1 2 1 5 arccos(0) cos − 2 arccos 2 arccos − 4 1 In order to have an inverse for tangent, we restrict the domain of tangent to the interval (−π=2; π=2). tan−1 x = y $ tan y = x y = tan x y = arctan x In other words, tan−1 x or arctan x is the ANGLE in the interval (−π=2; π=2) whose tangent is x. p arctan(1) tan−1(− 3) When combining trig and inverse trig, remember that an inverse trig expression is an ANGLE!! −1 4 −1 2 tan(sin 5 ) sin cos − 3 sin(tan−1 x) cot(cos−1 x) 2 2.2 The Limit of a Function x Introductory Example: Consider the function f(x) = p . The table below gives values of the x + 4 − 2 function when x is near 0. x f(x) x f(x) −0:5 3:8708287 0.5 4.1213203 −0:1 3:9748418 0:1 4.0248457 −0:05 3:9874607 0:05 4.0124612 −0:01 3:9974984 0:01 4.0024984 −0:001 3:999750 0:001 4.000250 −0:0001 3:999975 0:0001 4.000025 As the values of x get closer to 0 from both sides, the values of f(x) are getting closer to 4. This is written x mathematically as: lim p = 4 x!0 x + 4 − 2 lim f(x) = L x!a is read as \the limit as x approaches a of f(x) is L." This means the values of f(x) get closer and closer to L (or possibly equal to L) by letting x get closer and closer to a, but not equal to a. f(0) is undefined in the above example since f(0) = p 0 = 0 . When dealing with limits, we are 0+4−2 0 examining values as x approaches a, but not equal to a. [Note: We'll see later that the value of the function at a may or may not equal the limit.] In all three of these pictures, lim f(x) = 2. It doesn't matter what f(4) actually is or even if it exists. x!4 Left-Handed Limit: lim f(x) is the limit when ONLY looking at values of x approaching from the left, x!a− i.e. values of x less than a. Right-Handed Limit: lim f(x) is the limit ONLY looking at values of x approaching from the right, i.e. x!a+ values of x greather than a. The limit exists if and only if the left-handed and right-handed limits both exist and are equal. lim f(x) = L if and only if lim f(x) = L and lim f(x) = L x!a x!a− x!a+ 3 1 Infinite Limits: Calculate lim x!0 x2 x f(x) x f(x) −0:5 4 0:5 4 −0:1 100 0:1 100 −0:05 400 0:05 400 −0:01 10; 000 0:01 10; 000 −0:001 1; 000; 000 0:001 1; 000; 000 −0:0001 100; 000; 000 0:0001 100; 000; 000 As the values of x approach 0 from both sides, f(x) gets larger and larger without bound. In this case, we 1 say that lim = 1. x!0 x2 1 1 1 Example: Calculate lim , lim , and lim x!0+ x x!0− x x!0 x Definition: The line x = a is a vertical asymptote if the limit from the left, right, or both is 1 or −∞. Example: Consider the graph of f below. Find the indicated limits. lim f(x) lim f(x) lim f(x) x!4− x!4+ x!4 8 lim f(x) lim f(x) lim f(x) − + x!2 x!2 x!2 6 lim f(x) 4 x→−1 2 lim f(x) x→−6 −8 −6 −4 −2 2 4 6 8 lim f(x) lim f(x) lim f(x) x→−3− x→−3+ x→−3 −2 lim f(x) −4 x!6 −6 What are the vertical asymptotes of f(x)? −8 4 nonzero When the limit of a function at x = a is of the form 0 , then there is a vertical asymptote at x = a and the limit will be either 1 or −∞. Examples: x − 1 x − 1 x − 1 lim lim lim x→−5− x + 5 x→−5+ x + 5 x→−5 x + 5 x + 3 x3 lim lim x!2− 2 − x x!3 (x − 6)(x − 3)2 lim ln x lim ln(x3 − 64) x!0+ x!4+ Vertical Asymptotes vs Holes Be careful with rational functions. Just because there is division by zero does NOT mean there is a vertical asymptote. If a factor cancels completely from the denominator, then there is a hole there, not a vertical asymptote, because the division by 0 is “fixed.” x − 2 Find all vertical asymptotes of the function f(x) = . x2 − 6x + 8 5 2.3 Calculating Limits Using the Limit Laws Limit Laws: Suppose lim f(x) and lim g(x) exist and that c is any constant. x!a x!a 1. lim(f(x) ± g(x)) = lim f(x) ± lim g(x) x!a x!a x!a 2. lim cf(x) = c lim f(x) x!a x!a 3. lim f(x)g(x) = lim f(x) lim g(x) x!a x!a x!a lim f(x) 4. lim f(x) = x!a provided that lim g(x) 6= 0 x!a g(x) lim g(x) x!a x!a n 5. lim(f(x))n = lim f(x) x!a x!a pn q 6. lim f(x) = n lim f(x). If n is even, then we must have that lim f(x) > 0 x!a x!a x!a If f is not a piecewise function and a is in the domain of f, then lim f(x) = f(a). In other words, if you x!a can evaluate the function at a and you don't get division by 0 or something undefined, then that value IS the limit! lim (3x2 + 5x + 1)4 x→−2 p(x2 − 4)f(x) Given that lim f(x) = 16, calculate lim x!3 x!3 f(x) + x + 2 6 0 When the limit is of the form 0 we say the limit is indeterminate. Two things could be happening. There is either a vertical asymptote or a hole in the graph at x = a. If after using algebra, the limit simplifies to nonzero 0 , then there is a vertical asymptote. If the limit simplifies to an actual number, then there is a hole. 0 So, if you get a limit of the form 0 , you must USE ALGEBRA to determine the limit. Some methods used are expanding, factoring, or multiplying by the conjugate of a radical. x2 − x − 12 lim x!4 x2 − 16 (h − 4)2 − 16 lim h!0 h t − 2 lim t!2 (t − 2)3 p x + 3 − 2 lim x!1 x − 1 7 For vector functions, if r(t) = hf(t); g(t)i, then lim r(t) = lim f(t); lim g(t) t!a t!a t!a provided the limits of the component functions exist. * + 9t−1 − 3(t − 2)−1 t − 3 Calculate lim r(t), where r(t) = ; p t!3 t − 3 t2 + 7 − 4 8 8 x if x < 0 > > x2 if 0 ≤ x < 2 <> Let f(x) = 8 − x if 2 ≤ x < 5 > −2 if x = 5 > :> x − 2 if x > 5 Calculate lim f(x), lim f(x), and lim f(x) or explain why the limit does not exist. x!0 x!2 x!5 Recall the definition of the absolute value function: ( ( jxj = jx + 4j = x2 + x Calculate lim or explain why the limit does not exist. x!0 jxj 9 jx − 3j Calculate lim or explain why the limit does not exist. x!3 6 − 2x Squeeze Theorem: If g(x) ≤ h(x) ≤ f(x) for all x in an interval that contains a (except possibly at a) and lim g(x) = lim f(x) = L x!a x!a then lim h(x) = L x!a Example: If 4x − 2 ≤ f(x) ≤ x2 + 2 for 0 ≤ x ≤ 3, find lim f(x). x!2 Example: Find lim x2 sin 1 . x!0 x 10 2.5 Continuity In Section 2.3 we saw that the limit as x approaches a can sometimes be found by evaluating the function at a. If this is the case, then the function is continuous.