Oswaldo J.

The point $ P(2, -1) $ lies on the curve $ y = 1/(1-x) $.

(a) If $ Q $ is the point $ (x, 1/(1-x)) $, use your calculator to find the slope of the secant line $ PQ $ (correct to six decimal places) for the following values of $ x $:
(i) $ 1.5 $ (ii) $ 1.9 $ (iii) $ 1.99 $ (iv) $ 1.999 $
(v) $ 2.5 $ (vi) $ 2.1 $ (vii) $ 2.01 $ (viii) $ 2.001 $

(b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at $ P(2, -1) $.

(c) Using the slope from part (b), find an equation of the tangent line to the curve at $ P(2, -1) $.

Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).

$ \displaystyle \lim_{h \to 0}\frac{(2+h)^5 - 32}{h} $,
$ h = \pm 0.5, \pm 0.1, \pm 0.01, \pm 0.001, \pm 0.0001 $

(a) Evaluate $ h(x) = (\tan x - x)/x^3 $ for $ x $ = 1, 0.5, 0.1, 0.05, 0.01, and 0.005.

(b) Guess the value of $ \displaystyle \lim_{x \to 0}\frac{\tan x - x}{x^3} $.

(c) Evaluate $ h(x) $ for successively smaller values of $ x $ until you finally reach a value of 0 for $ h(x) $. Are you still confident that your guess in part (b) is correct? Explain why you eventually obtained 0 values. (In Section 4.4 a method of evaluating this limit will be explained.)

(d) Graph the function $ h $ in the viewing rectangle $ [-1, 1] $ by $ [0, 1] $. Then zoom in toward the point where the graph crosses the $ y $ -axis to estimate the limit of $ h(x) $ as $ x $ approaches 0. Continue to zoom in until you observe distortions in the graph of $ h $. Compare with the results of part (c).

Let $ g(x) = \text{sgn}(\sin x) $.

(a) Find each of the following limits or explain why it does not exist.
(i) $ \displaystyle \lim_{x \to 0^+}g(x) $
(ii) $ \displaystyle \lim_{x \to 0^-}g(x) $
(iii) $ \displaystyle \lim_{x \to 0}g(x) $
(iv) $ \displaystyle \lim_{x \to \pi^+}g(x) $
(v) $ \displaystyle \lim_{x \to \pi^-}g(x) $
(vi) $ \displaystyle \lim_{x \to \pi}g(x) $

(b) For which values of $ a $ does $ \displaystyle \lim_{x \to a}g(x) $ not exist?
(c) Sketch a graph of $ g $.

(a) Find a number $ \delta $ such that if $ | x - 2 | < \delta $, then $ | 4x - 8 | < \varepsilon $, where $ \varepsilon = 0.1 $.

(b) Repeat part (a) with $ \varepsilon = 0.01 $.

Prove the statement using the $ \varepsilon $, $ \delta $ definition of a limit.

$ \displaystyle \lim_{x \to 4}\frac{x^2 - 2x - 8}{x - 4} = 6 $