Advertisements
Advertisements
Question
Choose the correct alternative:
`lim_(x -> 3) [x]` =
Options
2
3
does not exist
0
Advertisements
Solution
does not exist
APPEARS IN
RELATED QUESTIONS
Evaluate the following limit:
`lim_(y -> -3) [(y^5 + 243)/(y^3 + 27)]`
Evaluate the following limit:
`lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following limit :
`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> 2)(2x + 3)` = 7
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> 2) (x^2 - 1)` = 3
Evaluate the following :
Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(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 -> 2) f(x)` where `f(x) = {{:(4 - x",", x ≠ 2),(0",", x = 2):}`
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)`
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Evaluate the following limits:
`lim_("h" -> 0) (sqrt(x + "h") - sqrt(x))/"h", x > 0`
Evaluate the following limits:
`lim_(x -> oo) (x^3 + x)/(x^4 - 3x^2 + 1)`
Evaluate the following limits:
`lim_(x -> oo) (x^4 - 5x)/(x^2 - 3x + 1)`
Evaluate the following limits:
`lim_(x -> oo) (1 + x - 3x^3)/(1 + x^2 +3x^3)`
Evaluate the following limits:
`lim_(x -> oo)(1 + "k"/x)^("m"/x)`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/(sin 5x)`
Evaluate the following limits:
`lim_(x -> pi) (1 + sinx)^(2"cosec"x)`
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.
