Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x -> 0) (2 "arc"sinx)/(3x)`
Advertisements
उत्तर
We know `lim_(x -> 0) (sin^-1 x)/x` = 1
`lim_(x -> 0) (2"arc" sinx)/(3x) = lim_(x -> 0) 2/3 (sin^-1 x)/x`
= `2/3 lim_(x -> 0) (sin^-1 x)/x`
`lim_(x -> 0) (2"arc" sinx)/(3x) = 2/3 xx 1`
= `2/3`
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 -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
Evaluate the following limit :
If `lim_(x -> 5) [(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> 1) (x^2 + x + 1)` = 3
Evaluate the following :
`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(x - 1)]`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) sin x/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) | 0.99833 | 0.99998 | 0.99999 | 0.99999 | 0.99998 | 0.99833 |
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 -> 1) sin pi x`
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
Sketch the graph of a function f that satisfies the given value:
f(– 2) = 0
f(2) = 0
`lim_(x -> 2) f(x)` = 0
`lim_(x -> 2) f(x)` does not exist.
Evaluate the following limits:
`lim_(x -> oo) (1 + x - 3x^3)/(1 + x^2 +3x^3)`
Evaluate the following limits:
`lim_(x -> 0) (sin^3(x/2))/x^2`
Evaluate the following limits:
`lim_(x -> 0) (sin("a" + x) - sin("a" - x))/x`
Evaluate the following limits:
`lim_(x-> 0) (1 - cos x)/x^2`
Evaluate the following limits:
`lim_(x -> oo) x [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)]`
Evaluate the following limits:
`lim_(x -> pi) (sin3x)/(sin2x)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(2) - sqrt(1 + cosx))/(sin^2x)`
Choose the correct alternative:
`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is
Choose the correct alternative:
The value of `lim_(x -> 0) sinx/sqrt(x^2)` is
`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.
`lim_(x→∞)((x + 7)/(x + 2))^(x + 4)` is ______.
