abannavong 3 years ago I need AP Calculus help!!

1. abannavong

i will post the assignment soon!

2. abannavong

Help with all plz!

3. abannavong

4. jldstuff393

hi

5. abannavong

the circled questions btw

6. jldstuff393

for the first one, its just 8 thats posted?

7. jldstuff393

i mean circled

8. abannavong

its numbers 10, 14, 16, 18, 22

9. jldstuff393

oh i was looking at the first one

10. abannavong

kk

11. jldstuff393

oh man, that second one's really hard to read lol so lets do the first one

12. abannavong

kk

13. jldstuff393

k so for 8 what is x approaching? i cant see

14. abannavong

we dont have to do number 8 lol

15. jldstuff393

:\ it looked circled to me lol

16. abannavong

its only numbers 10, 14, 16, 18, 22

17. jldstuff393

do you not have to do the problem in your first attachment?

18. abannavong

oh yeah duh i forgot i do lolz

19. jldstuff393

haha ok cool well i basically cant read any of them cuz they're blurry so can you just type out one that you need help with?

20. abannavong

kk

21. abannavong

hang on

22. abannavong

number 8 its as x goes to infinity

23. jldstuff393

so it's as x-> infinity of (2x^2)/(x^2 +3) ? is that right?

24. abannavong

2x^5/ x^5+3

25. jldstuff393

oh ok.

26. jldstuff393

have y'all talked about L'Hospital's Rule yet?

27. abannavong

i dont think so

28. jldstuff393

k.

29. abannavong

what is the L'hospitals rule? can u plz explain

30. jldstuff393

i don't think you're supposed to use it if you haven't covered it yet. you don't need it anyway.

31. jldstuff393

so you start by considering the terms with the highest exponents

32. jldstuff393

if I remember correctly, since x is approaching infinity and it's a rational function, you can rewrite it only including the x's with the greatest exponents

33. jldstuff393

so what would you have if you did that?

34. abannavong

hang on im kinda confused

35. jldstuff393

that's ok. so what i'm saying is your next step is to rewrite the limit excluding the +3 in the denominator

36. abannavong

to exclude the +3 in the denominator how do u do tht?

37. jldstuff393

you can do this because when x goes to infinity, 2x^5 and x^5 approach infinity and the 3 basically doesn't change the fraction anymore

38. jldstuff393

because if you imagine really high numbers, say x=100, x^5 would be 10000000000 and 2x^5 would be 20000000000 and the +3 in the denominator makes less and less of a difference in the fraction. (10000000000/20000000003 is really close to 10000000000/20000000000)

39. abannavong

oh ok

40. jldstuff393

in other words, the fraction approaches 1/2

41. jldstuff393

im sorry, it approaches 2

42. jldstuff393

because 2x^5 is in the numerator

43. abannavong

oh ok so ur limit is basically 2 then right?

44. jldstuff393

yup

45. abannavong

oh ok now i get number 8

46. jldstuff393

so in questions like that, where x is approaching infinity, you can disregard any terms that arent the variables with the highest exponent. like if you had an x^4 being added/subtracted somewhere in the fraction, you could disregard that, too.

47. abannavong

why can u?

48. jldstuff393

for the same reason, because as x approaches infinity, they change the fraction less and less, even though it has x in it....those terms approach infinity slower than terms of higher powers.

49. abannavong

im kinda confused on tht part

50. jldstuff393

it's kind of counter-intuitive at first

51. abannavong

oh ok

52. jldstuff393

as x gets greater, the difference between x^5 and x^4 also gets greater. if x is 10, then x^5 is 100000 and x^4 is 10000, x^5 is 10 times greater than x^4. but when x= 100, x^5 becomes 100 times bigger than x^4, and so on, until x^5 is infinitely bigger than x^4 and you can disregard x^4.

53. abannavong

oh ok now i get it

54. jldstuff393

:) cool

55. abannavong

can u help me with the continuity limits and one-sided ones those ones are the ones that r confusing

56. jldstuff393

i can help w one more then i need to get my own work done haha

57. abannavong

kk

58. jldstuff393

so i guess pick one for us to do

59. abannavong

kk

60. abannavong

i think i get the rest on my THQ but i need help with the continuity ones

61. jldstuff393

k

62. abannavong

let me post the problem

63. jldstuff393

kk

64. abannavong

$\lim_{x \rightarrow 4-} \frac{ \sqrt{x}-2 }{ x-4 }$

65. jldstuff393

so first notice that when x = 4, f(x) = 0/0, which is undefined.

66. abannavong

yeah

67. jldstuff393

and since x is approaching 4 from the left (less than 4, getting greater) both the numerator and denominator are going to be negative...do u get that?

68. abannavong

can u explain it a bit more im still a bit confused

69. jldstuff393

sure. so the way i think of it, is to basically imagine x as some number a little less than 4.. 3.something. when you plug that number into the denominator, 3.something - 4 is always going to be negative. (anything less than 4) - 4 will be negative, right? (3-4=-1, 3.5-4=-0.5, 3.99-4 = -0.01)

70. abannavong

oh yeah so its like it keeps on approaching 4 right?

71. jldstuff393

and for the numerator, you can imagine that since the square root of 4 is 2, the square root of anything less than 4 will be less than 2... when you subtract 2 from that number, the result is negative

72. jldstuff393

ya it keeps approaching 4

73. jldstuff393

but from the less-than-4 side because it has the minus sign (x->4-)

74. abannavong

basically the limit for it will negative right?

75. jldstuff393

well the numerator and denominator are both negative, right?

76. jldstuff393

so the fraction comes out positive

77. abannavong

oh ok

78. abannavong

why r limits soo confusing :3

79. jldstuff393

lol i know. wanna see my limit problem that no one will answer?

80. jldstuff393

Limit as x->(pi/2) of [tanx+1/(x-pi/2)]?

81. jldstuff393

thats my question that no one will answer :(

82. jldstuff393

i had it on a quiz today

83. abannavong

can u do the substitution method for tht one

84. abannavong

i think

85. abannavong

idk

86. jldstuff393

no because tan of pi/2 is undefined, it goes to neg infinity and infinity from the right and left....

87. abannavong

oh yeah ur right oops lol

88. jldstuff393

ya...it's brutal.

89. jldstuff393

anyway

90. abannavong

but im still kinda confused on the continuity limits and one-sided ones

91. nj1202

oh boy satellite's here.... it's about to get real xD

92. jldstuff393

i just realized that your problem is tricky. you need to factor your x-4 into (sqrt(x)-2)(sqrt(x)+2) then cancel out the numerator

93. satellite73

$\lim_{ x\to\frac{\pi}{2}}\frac{\tan(x)+1}{x-\frac{\pi}{2}}$?

94. jldstuff393

oooh well that's my problem but not hers

95. jldstuff393

mines posted

96. satellite73

oh ok

97. jldstuff393

somewhere down the list!

98. abannavong

@jldstuff393 how do u do my problem kinda confuzzled

99. jldstuff393

i know i know. u remember difference of squares? like (x^2 - 4) = (x-2)(x+2)?

100. satellite73

is this the problem $\lim_{x \rightarrow 4-} \frac{ \sqrt{x}-2 }{ x-4 }$?

101. abannavong

oh yeah i remember different of squares

102. abannavong

yeah @satellite73 thts my problem

103. jldstuff393

so unless satellite has a less tricky way to solve it, you can factor (x-4) as a difference of squares... (sqrt(x)-2)(sqrt(x)+2)

104. jldstuff393

does that make sense?

105. satellite73

the gimmick here is to "rationalize the numerator" by multiplying by the conjugate of $$\sqrt{x}-2$$ which is $$\sqrt{x}+2$$

106. jldstuff393

cuz when u have that, you can cancel out the (sqrt(x)-2) in the numerator and denominator

107. abannavong

so u multpily the top and bottom by $\sqrt{x} +2$

108. satellite73

this works because $$(a+b)(a-b)=a^2-b^2$$ and so $$(\sqrt{x}-2)(\sqrt{x}+2)=x-4$$

109. satellite73

that cancels with the $$x-4$$ in the denominator, which is what you want, you want to cancel the zero

110. abannavong

and x-4 can cancel in denominator

111. abannavong

oh ok!

112. abannavong

so ur limit is one right?

113. satellite73

leaves you with $\lim_{x \rightarrow 4} \frac{ \sqrt{x}-2 }{ x-4 }=\lim_{x \rightarrow 4} \frac{1 }{ \sqrt{x}+2}$\]

114. abannavong

oh ok i get

115. satellite73

no limit is not one, now you have to replace $$x$$ by 4

116. abannavong

once u replace x with 4 u get 1/ 4 right?

117. satellite73

yup

118. satellite73

whenever you get $$\frac{0}{0}$$ you need some gimmick to factor and cancel to get rid of the zeros

119. jldstuff393

satellite would you mind helping me after you're done here? O.o good luck guys!

120. abannavong

oh ok

121. satellite73

sometime sit is easy sometimes it is a pain. but you get used to it

122. abannavong

@satellite73 i have some more questions!

123. satellite73

@jldstuff393 sure if i can what is the question

124. satellite73

@abannavong go ahead and post them, i will look

125. jldstuff393

i'll repost it, thanks!

126. abannavong

kk! thanks @satellite73

127. abannavong

14, 16, and 18

128. satellite73

i can't really read it, but for 16 replace $$x$$ by 2 in both formulas if you get the same number, that is the limit if you get different numbers, there is no limit

129. abannavong

number 14 its as deltax goes to 0 with the little plus sign

130. satellite73

i can't read 18 either is it $$\lim_{ x\to 1^+}$$ or $$\lim_{x\to 1^-}$$?

131. abannavong

x-> 1+

132. satellite73

then replace $$x$$ by 1 in the second formula you get $$1-1=0$$

133. abannavong

oh ok

134. abannavong

can u help with 14 also

135. abannavong

its confusing

136. satellite73

sure, but i can't really read it can you post in a new question? you do not need to write $$\Delta x$$ you can just write $$h$$ instead

137. satellite73

it is actually very straight forward we will do a little algebra and that is all

138. abannavong

kk hang on dinner quickly

139. satellite73

$(x+h)^2+x+h-(x^2+x)=x^2+2xh+h^2+x+h-x^2-x$ $\cancel{x^2}+2xh+h^2+\cancel{x}+h-\cancel{x^2}-\cancel{x}=2xh+h+h^2$ divide by $$h$$ and get $$2x+1+h$$ let $$h\to 0$$ be replacing $$h$$ by 0 and you get $$2x+1$$

140. abannavong

oh ok! @satellite73 im back from dinner

141. satellite73

that was a quick meal, you should take your time i wrote the solution above, but i used $$h$$ instead of $$\Delta x$$ it is mostly just algebra, to get rid of the factor of $$\Delta x$$ in the denominator, then you replace it by 0

142. abannavong

i did eat dinner

143. abannavong

im eating and doing work lolz

144. satellite73

you are either eating late or are on the left coast

145. abannavong

im on the west coast lolz

146. satellite73

after you finish eating, read my solution above i hope all the steps are clear they are almost all algebra steps, only at the end do you replace $$h$$ by 0

147. abannavong

kk

148. abannavong

oh ok its jst basic algebra right? so u foil the (x+h)^2 right?

149. satellite73

yes exactly it is $$(x+h)(x+h)=x^2+2xh+h^2$$

150. abannavong

oh ok and then from there u can combine the like terms and get 2xh+h+h^2

151. satellite73

yes,

152. satellite73

that is all in the numerator you get $\frac{2xh+h+h^2}{h}$

153. abannavong

and then can u factor out the h on the numerator

154. satellite73

yes factor and cancel or divide each term by $$h$$ it is the same thing

155. abannavong

oh ok and then u get 2x+1

156. satellite73

yes

157. abannavong

oh ok! i get tht now

158. satellite73

whew that was exhausting

159. abannavong

$\lim_{x \rightarrow2 ^{+}}(2x-\left[ \left| x \right| \right]$ help im confused with this one also

160. satellite73

is that an absolute value sign?

161. satellite73

or is that the floor function?

162. abannavong

its like a bracket and then two straight lines inside

163. satellite73

do you know what that function is called?

164. abannavong

ughhh i dont remember what it was called

165. satellite73

it is either called "floor" or "greatest integer"

166. abannavong

is it kinda like a piecewise function or something like tht

167. abannavong

oh yeah i remember!

168. satellite73

so if $$x>2$$ but close to 2, then $$[x]=2$$

169. satellite73

so you have $\lim_{x\to 2}2x-2=4-2=2$

170. satellite73

to be more precise $\lim_{x\to 2^+}2x-2=4-2=2$

171. abannavong

oh ok! at first it seemed confusing but oh ok now i get it!

172. satellite73

whereas $\lim_{x\to 2^-}2x-2=4-1=3$ because if $$x<2$$ but close to 2, then $$[x]=1$$

173. abannavong

oh ok! tht seems soo much easier now!

174. satellite73

yes if you know what it means it is not hard here is a nice picture of $$y=2x-[x]$$ http://www.wolframalpha.com/input/?i=2x-floor%28x%29

175. abannavong

oh ok! i get it! i have another question again hang on

176. abannavong

so it says find the x-values(if any) at which f is not continuous. which of the discontinuities are removable. f(x)= 1/ x^2+1

177. satellite73

$f(x)=\frac{1}{x^2+1}$ has no discontinuities, since it is a rational function and the denominator is never 0 since $$x^2\geq 0$$ for any $$x$$ you can see that $$x^2+1\geq 1$$ no matter what $$x$$ is, so this function has no discontinuities. as the denominator is never zero

178. abannavong

oh ok thank you @satellite73

179. satellite73

you are welcome enough math, study something else or do something more constructive

180. abannavong

pretty much tht was my last problem! time to find a prayer for theology class! thank you @satellite73 and study for english test tomorrow on grapes of wrath

181. satellite73