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x + 5

Lim f (x) = ------- = 4

x-> -1 2x+3

using the definition.

you have to prove that, given an__epsilon ∈ __,__ there exists a__ __delta d__ such that

__| f(x) - 4 | < ∈ whenever | x - (-1) | < d __that is | x + 1 | < d.

So let f(x) - 4 < ∈ = > f(x) < 4 + ∈

(x+5)/(2x+3) < 4+∈ => x+5 < (8+2∈)x +12 + 3∈

=> x (-7-2∈) < 7 + 3∈

=> x < - (7 +3∈) / (7+2∈)

=> x+1 < 1 - (7 +3∈) / (7+2∈)

=> x+1 < -∈/7+2∈ this is equal to d.

since there exists a 'd' for every ∈. Limit exists and it is 4.

Lim f (x) = ------- = 4

x-> -1 2x+3

using the definition.

you have to prove that, given an

So let f(x) - 4 < ∈ = > f(x) < 4 + ∈

(x+5)/(2x+3) < 4+∈ => x+5 < (8+2∈)x +12 + 3∈

=> x (-7-2∈) < 7 + 3∈

=> x < - (7 +3∈) / (7+2∈)

=> x+1 < 1 - (7 +3∈) / (7+2∈)

=> x+1 < -∈/7+2∈ this is equal to d.

since there exists a 'd' for every ∈. Limit exists and it is 4.