I'm stuck and don't understand continuity and graphing these function limits. Would anyone be able to please explain it to me?

Q1:

The graph of y = i/4x _ x2 is given. Use transformations to create a function whose graph is as shown. Sketch the graph of the function. 4+x ifxa Y EXAMPLE 11 Show that the following limit is true. lim x3 sin(l) x -> D X SOLUTION First note that we cannot use x lim x3 sin (A) = Iim Xs- lim sin(l) x > D X x ) 0 x ) D X because the limit as x approaches 0 of sin(1/x) does not exist (see this example). Instead we apply the Squeeze Theorem, and so we need to nd a function fsmaller than g(x) = x5 sin(1/x) and a function h bigger than 9 such that both f(x) and h(x) approach 0. To do this we use our knowledge of the sine function. Because the sine of any number lies between and , we can write :| ssin(l) S 7. X Video Example4>)) Any inequality remains true when multiplied by a positive number. We know that X8 2 0 for all x and so, multiplying each side of inequalities by x5, we get s x\"3 sin(%) 5 as illustrated by the figure. We know that lim x8: and lim (x3)= . x > D x v 0 Taking f(x) = x3, g(x) = x3 sin(1/x), and h(x) = x8 in the Squeeze Theorem, we obtain lim x8 sin(l) = o. x > D X Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval. x)=x+ x4, [4,00) For a 4, we have Iim f(x) = lim (x + W) xra xla lim x+ lim ( ) xxa xra X'a xra xra l E || .1 A \ / Therefore, fis continuous at x = a for every 5 in (4, co). Also, lim f(x) = = r14), so fis continuous from the right at 4. Thus, fis continuous on [4, co). xut'r