## anonymous 3 years ago I need AP Calculus help!!

1. anonymous

i will post the assignment soon!

2. anonymous

Help with all plz!

3. anonymous

4. anonymous

hi

5. anonymous

the circled questions btw

6. anonymous

for the first one, its just 8 thats posted?

7. anonymous

i mean circled

8. anonymous

its numbers 10, 14, 16, 18, 22

9. anonymous

oh i was looking at the first one

10. anonymous

kk

11. anonymous

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

12. anonymous

kk

13. anonymous

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

14. anonymous

we dont have to do number 8 lol

15. anonymous

:\ it looked circled to me lol

16. anonymous

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

17. anonymous

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

18. anonymous

oh yeah duh i forgot i do lolz

19. anonymous

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. anonymous

kk

21. anonymous

hang on

22. anonymous

number 8 its as x goes to infinity

23. anonymous

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

24. anonymous

2x^5/ x^5+3

25. anonymous

oh ok.

26. anonymous

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

27. anonymous

i dont think so

28. anonymous

k.

29. anonymous

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

30. anonymous

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

31. anonymous

so you start by considering the terms with the highest exponents

32. anonymous

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. anonymous

so what would you have if you did that?

34. anonymous

hang on im kinda confused

35. anonymous

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

36. anonymous

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

37. anonymous

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. anonymous

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. anonymous

oh ok

40. anonymous

in other words, the fraction approaches 1/2

41. anonymous

im sorry, it approaches 2

42. anonymous

because 2x^5 is in the numerator

43. anonymous

oh ok so ur limit is basically 2 then right?

44. anonymous

yup

45. anonymous

oh ok now i get number 8

46. anonymous

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. anonymous

why can u?

48. anonymous

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. anonymous

im kinda confused on tht part

50. anonymous

it's kind of counter-intuitive at first

51. anonymous

oh ok

52. anonymous

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. anonymous

oh ok now i get it

54. anonymous

:) cool

55. anonymous

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

56. anonymous

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

57. anonymous

kk

58. anonymous

so i guess pick one for us to do

59. anonymous

kk

60. anonymous

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

61. anonymous

k

62. anonymous

let me post the problem

63. anonymous

kk

64. anonymous

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

65. anonymous

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

66. anonymous

yeah

67. anonymous

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. anonymous

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

69. anonymous

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. anonymous

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

71. anonymous

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. anonymous

ya it keeps approaching 4

73. anonymous

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

74. anonymous

basically the limit for it will negative right?

75. anonymous

well the numerator and denominator are both negative, right?

76. anonymous

so the fraction comes out positive

77. anonymous

oh ok

78. anonymous

why r limits soo confusing :3

79. anonymous

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

80. anonymous

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

81. anonymous

thats my question that no one will answer :(

82. anonymous

i had it on a quiz today

83. anonymous

can u do the substitution method for tht one

84. anonymous

i think

85. anonymous

idk

86. anonymous

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

87. anonymous

oh yeah ur right oops lol

88. anonymous

ya...it's brutal.

89. anonymous

anyway

90. anonymous

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

91. anonymous

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

92. anonymous

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. anonymous

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

94. anonymous

oooh well that's my problem but not hers

95. anonymous

mines posted

96. anonymous

oh ok

97. anonymous

somewhere down the list!

98. anonymous

@jldstuff393 how do u do my problem kinda confuzzled

99. anonymous

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

100. anonymous

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

101. anonymous

oh yeah i remember different of squares

102. anonymous

yeah @satellite73 thts my problem

103. anonymous

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. anonymous

does that make sense?

105. anonymous

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

106. anonymous

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

107. anonymous

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

108. anonymous

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

109. anonymous

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

110. anonymous

and x-4 can cancel in denominator

111. anonymous

oh ok!

112. anonymous

so ur limit is one right?

113. anonymous

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

114. anonymous

oh ok i get

115. anonymous

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

116. anonymous

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

117. anonymous

yup

118. anonymous

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

119. anonymous

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

120. anonymous

oh ok

121. anonymous

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

122. anonymous

@satellite73 i have some more questions!

123. anonymous

@jldstuff393 sure if i can what is the question

124. anonymous

@abannavong go ahead and post them, i will look

125. anonymous

i'll repost it, thanks!

126. anonymous

kk! thanks @satellite73

127. anonymous

14, 16, and 18

128. anonymous

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. anonymous

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

130. anonymous

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

131. anonymous

x-> 1+

132. anonymous

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

133. anonymous

oh ok

134. anonymous

can u help with 14 also

135. anonymous

its confusing

136. anonymous

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. anonymous

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

138. anonymous

kk hang on dinner quickly

139. anonymous

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

140. anonymous

oh ok! @satellite73 im back from dinner

141. anonymous

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. anonymous

i did eat dinner

143. anonymous

im eating and doing work lolz

144. anonymous

you are either eating late or are on the left coast

145. anonymous

im on the west coast lolz

146. anonymous

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. anonymous

kk

148. anonymous

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

149. anonymous

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

150. anonymous

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

151. anonymous

yes,

152. anonymous

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

153. anonymous

and then can u factor out the h on the numerator

154. anonymous

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

155. anonymous

oh ok and then u get 2x+1

156. anonymous

yes

157. anonymous

oh ok! i get tht now

158. anonymous

whew that was exhausting

159. anonymous

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

160. anonymous

is that an absolute value sign?

161. anonymous

or is that the floor function?

162. anonymous

its like a bracket and then two straight lines inside

163. anonymous

do you know what that function is called?

164. anonymous

ughhh i dont remember what it was called

165. anonymous

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

166. anonymous

is it kinda like a piecewise function or something like tht

167. anonymous

oh yeah i remember!

168. anonymous

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

169. anonymous

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

170. anonymous

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

171. anonymous

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

172. anonymous

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

173. anonymous

oh ok! tht seems soo much easier now!

174. anonymous

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. anonymous

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

176. anonymous

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. anonymous

$f(x)=\frac{1}{x^2+1}$ 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. anonymous

oh ok thank you @satellite73

179. anonymous

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

180. anonymous

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. anonymous