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 -> 3) 1/(x - 3)`
Advertisements
उत्तर
To find `lim_(x -> 3) 1/(x - 3)`
y = f(x) = `1/(x - 3)`
From the graph the value of the function at x = 3 the curve does not meet the line x = 3
∴ The value of the function is not defined at the point x = 3.
Hence `lim_(x -> 3) 1/(x - 3)` does not exist at x = 3.
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(x -> 3)[sqrt(2x + 6)/x]`
Evaluate the following limit :
`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
Evaluate the following :
Find the limit of the function, if it exists, at x = 1
f(x) = `{(7 - 4x, "for", x < 1),(x^2 + 2, "for", x ≥ 1):}`
Evaluate the following :
`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(x - 1)]`
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)`
Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.
f(x) = `{{:(sin x",", x < 0),(1 - cos x",", 0 ≤ x ≤ pi),(cos x",", x > pi):}`
Sketch the graph of a function f that satisfies the given value:
f(0) is undefined
`lim_(x -> 0) f(x)` = 4
f(2) = 6
`lim_(x -> 2) f(x)` = 3
Evaluate the following limits:
`lim_(x - 0) (sqrt(1 + x^2) - 1)/x`
Find the left and right limits of f(x) = `(x^2 - 4)/((x^2 + 4x+ 4)(x + 3))` at x = – 2
Evaluate the following limits:
`lim_(x -> 0)(1 + x)^(1/(3x))`
Evaluate the following limits:
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`
Evaluate the following limits:
`lim_(x -> 0) (sin("a" + x) - sin("a" - x))/x`
Evaluate the following limits:
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`
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:
If `f(x) = x(- 1)^([1/x])`, x ≤ 0, then the value of `lim_(x -> 0) f(x)` is equal to
Choose the correct alternative:
`lim_(x -> 3) [x]` =
Choose the correct alternative:
The value of `lim_(x -> 0) sinx/sqrt(x^2)` is
`lim_(x -> 5) |x - 5|/(x - 5)` = ______.
If f(x) is differentiable at x = 1 and `lim_(h → 0) 1/h f(1 + h) = 5`, then f' (1) is equal to ______.
