If lim f(x) and lim f(x) exist but are not equal, then x = c is a(n) jump discontinuity. X-C- XC+ If one or both of lim f(x) and lim f(x) is infinite, then x = c is a(n) removablex discontinuity even if f(x) itself is X-C- X- C+ not defined at x = c. 3 To determine the type of discontinuity at x = 3 find the one-sided limits as x approaches 2 x + 1 lim x- 3/2 - 6x - 9 X + 1 lim X3/2 + 6x - 9 State the type of discontinuity at x = infinite O removable jump none of these

Q: 1. Let f(x)= x²+x-6 |x2| 7 x b 2 x + a 3 2 if x 2

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Q: The graph below is the function f(x) O Q -5 -4 -3 -2 -1 x→-2+ 5+ 4 3 2 x -1 -2 -3 -4 Determine the…

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Q: Prove that limxS2 ƒ(x) = 4 if

A: Given that : The function is f(x) =x2,    x≠21  ,    x = 2 .

Q: Show directly from definition that if lim,l- f(x) = 2 and lim,1- 9(z) = 3, then lim,1- f(r) + g(x) =…

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Q: Find lim f(x) if, for all x > 1, бех 19 < f(x) < - 2ex VX - 1

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Q: (2) Use the definition of limit of a function to find lim x3+5x 2-22 - - 2 3x+2

A: Hence,By using the L'Hospital rule concept, we conclude that a given function limit does not…

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Q: x² – 9 (b) lim x→3- x + 3

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Q: Suppose lim f(x) = = x→2 1/2 4 -5, find lim x→2 x²| ƒ(x) + 6 x - 1

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Q: Show that lim(x+¹)=e².

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Q: Show by example that lim[f(x) +g(x)] can exist even if lim f(x) and lim g(x) do not exist.

A: Let,  f(x)=1x-c, g(x)=1c-x

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Q: if x1 then lim,-1 f(x) is 3 if x 1 (b) 3 3. Let f(x) = (a) 1 (c) not exist (d) other

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Q: x² +x-5 (1) Find lim 1-2x-xr

A: topic- limits

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Q: If f is C2 on an interval prove that f(2 + k) – 2/(2) + {(z =A) = r"(2). - lim h-0 h2 = f"(x). The…

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Q: The graph below is the function f(x) 25 lim f(x) = x4-5 20 25 -20 -15 -10 -5 f(-5) = 15 10 5 -5 -10…

A: A graph of the function is given.The aim is to find the limit value  and the function value .

Q: Use the graph of f(x) = X + 2x - x-8) to determine lim f(x) and lim f(x). X→-4 X-2

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Q: In(2+x) lim, x+-1 X+1

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Q: f(x, b) – f(a, b) fe(a, b) = lim x - a OTrue False

A: Given a limit : fxa,b=limx→a f(x,b)-f(a,b)x-a We need to find it it's true or false. Let's begin.