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## abannavong Group Title I need AP Calculus help!! 2 years ago 2 years ago

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1. abannavong Group Title

i will post the assignment soon!

2. abannavong Group Title

Help with all plz!

3. abannavong Group Title

4. jldstuff393 Group Title

hi

5. abannavong Group Title

the circled questions btw

6. jldstuff393 Group Title

for the first one, its just 8 thats posted?

7. jldstuff393 Group Title

i mean circled

8. abannavong Group Title

its numbers 10, 14, 16, 18, 22

9. jldstuff393 Group Title

oh i was looking at the first one

10. abannavong Group Title

kk

11. jldstuff393 Group Title

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

12. abannavong Group Title

kk

13. jldstuff393 Group Title

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

14. abannavong Group Title

we dont have to do number 8 lol

15. jldstuff393 Group Title

:\ it looked circled to me lol

16. abannavong Group Title

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

17. jldstuff393 Group Title

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

18. abannavong Group Title

oh yeah duh i forgot i do lolz

19. jldstuff393 Group Title

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 Group Title

kk

21. abannavong Group Title

hang on

22. abannavong Group Title

number 8 its as x goes to infinity

23. jldstuff393 Group Title

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

24. abannavong Group Title

2x^5/ x^5+3

25. jldstuff393 Group Title

oh ok.

26. jldstuff393 Group Title

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

27. abannavong Group Title

i dont think so

28. jldstuff393 Group Title

k.

29. abannavong Group Title

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

30. jldstuff393 Group Title

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 Group Title

so you start by considering the terms with the highest exponents

32. jldstuff393 Group Title

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 Group Title

so what would you have if you did that?

34. abannavong Group Title

hang on im kinda confused

35. jldstuff393 Group Title

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 Group Title

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

37. jldstuff393 Group Title

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 Group Title

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 Group Title

oh ok

40. jldstuff393 Group Title

in other words, the fraction approaches 1/2

41. jldstuff393 Group Title

im sorry, it approaches 2

42. jldstuff393 Group Title

because 2x^5 is in the numerator

43. abannavong Group Title

oh ok so ur limit is basically 2 then right?

44. jldstuff393 Group Title

yup

45. abannavong Group Title

oh ok now i get number 8

46. jldstuff393 Group Title

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 Group Title

why can u?

48. jldstuff393 Group Title

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 Group Title

im kinda confused on tht part

50. jldstuff393 Group Title

it's kind of counter-intuitive at first

51. abannavong Group Title

oh ok

52. jldstuff393 Group Title

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 Group Title

oh ok now i get it

54. jldstuff393 Group Title

:) cool

55. abannavong Group Title

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

56. jldstuff393 Group Title

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

57. abannavong Group Title

kk

58. jldstuff393 Group Title

so i guess pick one for us to do

59. abannavong Group Title

kk

60. abannavong Group Title

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

61. jldstuff393 Group Title

k

62. abannavong Group Title

let me post the problem

63. jldstuff393 Group Title

kk

64. abannavong Group Title

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

65. jldstuff393 Group Title

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

66. abannavong Group Title

yeah

67. jldstuff393 Group Title

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 Group Title

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

69. jldstuff393 Group Title

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 Group Title

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

71. jldstuff393 Group Title

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 Group Title

ya it keeps approaching 4

73. jldstuff393 Group Title

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

74. abannavong Group Title

basically the limit for it will negative right?

75. jldstuff393 Group Title

well the numerator and denominator are both negative, right?

76. jldstuff393 Group Title

so the fraction comes out positive

77. abannavong Group Title

oh ok

78. abannavong Group Title

why r limits soo confusing :3

79. jldstuff393 Group Title

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

80. jldstuff393 Group Title

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

81. jldstuff393 Group Title

thats my question that no one will answer :(

82. jldstuff393 Group Title

i had it on a quiz today

83. abannavong Group Title

can u do the substitution method for tht one

84. abannavong Group Title

i think

85. abannavong Group Title

idk

86. jldstuff393 Group Title

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

87. abannavong Group Title

oh yeah ur right oops lol

88. jldstuff393 Group Title

ya...it's brutal.

89. jldstuff393 Group Title

anyway

90. abannavong Group Title

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

91. nj1202 Group Title

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

92. jldstuff393 Group Title

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 Group Title

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

94. jldstuff393 Group Title

oooh well that's my problem but not hers

95. jldstuff393 Group Title

mines posted

96. satellite73 Group Title

oh ok

97. jldstuff393 Group Title

somewhere down the list!

98. abannavong Group Title

@jldstuff393 how do u do my problem kinda confuzzled

99. jldstuff393 Group Title

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

100. satellite73 Group Title

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

101. abannavong Group Title

oh yeah i remember different of squares

102. abannavong Group Title

yeah @satellite73 thts my problem

103. jldstuff393 Group Title

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 Group Title

does that make sense?

105. satellite73 Group Title

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

106. jldstuff393 Group Title

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

107. abannavong Group Title

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

108. satellite73 Group Title

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

109. satellite73 Group Title

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

110. abannavong Group Title

and x-4 can cancel in denominator

111. abannavong Group Title

oh ok!

112. abannavong Group Title

so ur limit is one right?

113. satellite73 Group Title

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

114. abannavong Group Title

oh ok i get

115. satellite73 Group Title

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

116. abannavong Group Title

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

117. satellite73 Group Title

yup

118. satellite73 Group Title

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

119. jldstuff393 Group Title

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

120. abannavong Group Title

oh ok

121. satellite73 Group Title

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

122. abannavong Group Title

@satellite73 i have some more questions!

123. satellite73 Group Title

@jldstuff393 sure if i can what is the question

124. satellite73 Group Title

@abannavong go ahead and post them, i will look

125. jldstuff393 Group Title

i'll repost it, thanks!

126. abannavong Group Title

kk! thanks @satellite73

127. abannavong Group Title

14, 16, and 18

128. satellite73 Group Title

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 Group Title

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

130. satellite73 Group Title

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

131. abannavong Group Title

x-> 1+

132. satellite73 Group Title

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

133. abannavong Group Title

oh ok

134. abannavong Group Title

can u help with 14 also

135. abannavong Group Title

its confusing

136. satellite73 Group Title

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 Group Title

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

138. abannavong Group Title

kk hang on dinner quickly

139. satellite73 Group Title

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

140. abannavong Group Title

oh ok! @satellite73 im back from dinner

141. satellite73 Group Title

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 Group Title

i did eat dinner

143. abannavong Group Title

im eating and doing work lolz

144. satellite73 Group Title

you are either eating late or are on the left coast

145. abannavong Group Title

im on the west coast lolz

146. satellite73 Group Title

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 Group Title

kk

148. abannavong Group Title

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

149. satellite73 Group Title

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

150. abannavong Group Title

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

151. satellite73 Group Title

yes,

152. satellite73 Group Title

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

153. abannavong Group Title

and then can u factor out the h on the numerator

154. satellite73 Group Title

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

155. abannavong Group Title

oh ok and then u get 2x+1

156. satellite73 Group Title

yes

157. abannavong Group Title

oh ok! i get tht now

158. satellite73 Group Title

whew that was exhausting

159. abannavong Group Title

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

160. satellite73 Group Title

is that an absolute value sign?

161. satellite73 Group Title

or is that the floor function?

162. abannavong Group Title

its like a bracket and then two straight lines inside

163. satellite73 Group Title

do you know what that function is called?

164. abannavong Group Title

ughhh i dont remember what it was called

165. satellite73 Group Title

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

166. abannavong Group Title

is it kinda like a piecewise function or something like tht

167. abannavong Group Title

oh yeah i remember!

168. satellite73 Group Title

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

169. satellite73 Group Title

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

170. satellite73 Group Title

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

171. abannavong Group Title

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

172. satellite73 Group Title

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

173. abannavong Group Title

oh ok! tht seems soo much easier now!

174. satellite73 Group Title

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 Group Title

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

176. abannavong Group Title

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 Group Title

$f(x)=\frac{1}{x^2+1}$ 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 Group Title

oh ok thank you @satellite73

179. satellite73 Group Title

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

180. abannavong Group Title

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 Group Title

good luck with the joads

182. abannavong Group Title

lolz thanks i have to find the intercalary chapters to write about :3 i dont even know what those r