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:
`lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
Evaluate the following limit :
If `lim_(x -> 5) [(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
Evaluate the following limit :
`lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
Evaluate the following limit :
`lim_(x -> 1) [(x + x^3 + x^5 + ... + x^(2"n" - 1) - "n")/(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 -> 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 the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
Evaluate the following limits:
`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) ("a" > "b")`
Evaluate the following limits:
`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9))`
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)`
Show that `lim_("n" -> oo) (1 + 2 + 3 + ... + "n")/(3"n"^2 + 7n" + 2) = 1/6`
Evaluate the following limits:
`lim_(x-> 0) (1 - cos x)/x^2`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/x`
Evaluate the following limits:
`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`
Evaluate the following limits:
`lim_(x -> pi) (1 + sinx)^(2"cosec"x)`
Evaluate the following limits:
`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 -4x + 2))^x`
Choose the correct alternative:
`lim_(x -> oo) (1/"n"^2 + 2/"n"^2 + 3/"n"^2 + ... + "n"/"n"^2)` is
`lim_(x -> 5) |x - 5|/(x - 5)` = ______.
