Advertisements
Advertisements
प्रश्न
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 2) f(x)` where `f(x) = {{:(4 - x",", x ≠ 2),(0",", x = 2):}`
Advertisements
उत्तर
`f(x) = {{:(4 - x",", x ≠ 2),(0",", x = 2):}`
To find `lim_(x -> 2) f(x)`
From the figure the value of the function at x = 2 is y = f(2) = 2
∴ `lim_(x -> 2) f(x)` = 2
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
If `lim_(x -> 1)[(x^4 - 1)/(x - 1)]` = `lim_(x -> "a")[(x^3 - "a"^3)/(x - "a")]`, find all possible values of a
Evaluate the following limit :
`lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
Evaluate the following :
`lim_(x -> 0) [(sqrt(1 - cosx))/x]`
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 3) (4 - x)`
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 5) |x - 5|/(x - 5)`
Evaluate : `lim_(x -> 3) (x^2 - 9)/(x - 3)` if it exists by finding `f(3^-)` and `f(3^+)`
Evaluate the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
Evaluate the following limits:
`lim_(x -> 0)(1 + x)^(1/(3x))`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + "a"^2) - "a")/(sqrt(x^2 + "b"^2) - "b")`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 + sinx) - sqrt(1 - sinx))/tanx`
Evaluate the following limits:
`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 -4x + 2))^x`
Choose the correct alternative:
`lim_(x - pi/2) (2x - pi)/cos x`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
`lim_(x -> 0) (8^x - 4x - 2^x + 1^x)/x^2` =
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
Choose the correct alternative:
`lim_(x -> 0) ("e"^(sin x) - 1)/x` =
`lim_(x -> 5) |x - 5|/(x - 5)` = ______.
`lim_(x -> 0) (sin 4x + sin 2x)/(sin5x - sin3x)` = ______.
