Aevaluate the following limits evaluate the following limits...

a) Evaluate the following limits: 1. limx→−7^- 1/(x^2(x+7)) = 2. limx→0 1/(x^2(x+7)) = 3. limx→5 2/(x-5)^6 = 4. limx→3^- 2/(x-3) = b) Evaluate the following limits. (a) limx→∞ 2/(e^x+8) (b) limx→−∞ 2/(e^x+8) c) A function is said to have a horizontal asymptote if either the limit at infinity exists or the limit at negative infinity exists. Show that each of the following functions has a horizontal asymptote by calculating the given limit. limx→∞ -7x/(11+2x) limx→−∞ (5x-4)/(x^3+8x-15) limx→∞ (x^2-3x-2)/(10-9x^2) limx→∞ √(x^2+10x)/(9-6x) limx→−∞ √(x^2+10x)/(9-6x)

Transcript

00:02 First a part, we need to find the limit, left -hand limit at 7 of 1 over x square multiplied to x plus 7.

00:13 So this is equal to the limit is tending to 0, 1 over 7 negative edge to whole square multiplied to 7 negative as negative 7.

00:25 Simson said it can sell out, apply the limit here that is 1 over negative 0, that is a negative infinity.

00:35 Now in part 2 we need to find the limit x tending to 0 1 over x squared multiply to x plus 7.

00:43 This is equal to 1 over 0 multiplied to 7, that is positive infinity.

00:50 In part 3 we need to find the limit x tending to 5, 2 over x negative 5 to the power 6.

00:59 Is equal to 1 over 0 this is equal to positive infinity now in part 4 we'll find the left -hand limit at 3 of the given function 2 over x negative 3 is equal to the limit a standing to 0 2 over 3 negative as negative 3 if they get cancelled out it gave negative 2 over 0 that is negative infinity now in part b so in first a part we need to solve the limit x -tend to infinity 2 over 8 to the power x plus 8.

01:46 So it gives 2 over infinity that is 0.

01:51 So now in part 2 we need to find the limit x 10 to negative infinity of 2 over 8 to the of x plus 8 so that is 2 over 8 to the bar of negative between 8 to 0 plus 8 this equal to 1 over 4 so this is a required answer now in part c so in first a part we need to find the limit x standing infinity of negative 7x over 11 plus 2 times of x this is equal to the limit x tending to infinity divide the numerate denoted by x that is negative 7 over 11 over x plus 2 apply the limit here it gave negative 7 over 0 plus that is a negative 3 .5 so this is our horizontal asymptot now in part 2 you need to find the limit x 10 negative of 5x negative 4 over x q plus 8x negative 15 so this is equal to take the x as a common.

03:17 So that is limit x tangent negative infinity...

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