Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x -> ) (sinx(1 - cosx))/x^3`
Advertisements
उत्तर
We know `lim_(x -> 0) sinx/x` = 1
`lim_(x -> 0) (sinx(1 - cosx))/x^3 = lim_(x -> 0) (sinx xx 2 sin^2 x/2)/x^3`
= `lim_(x -> 0) (sinx/x) xx 2 (sin^2 x/2)/x^2`
= `lim_(x -> 0) (sinx/x) xx 2 (sin^2 x/2)/(2^2 xx x^2/2^2)`
= `lim_(x -> 0) [(sinx/x) xx 1/2 ((sin (x/2))/((x/2)))^2]`
= `lim_(x -> 0) (sinx/x) xx 1/2 (lim_(x/2 -> 0) (sin x/2)/(x/2))^2`
= `1 xx 1/2 xx 1`
`lim_(x -> ) (sinx(1 - cosx))/x^3 = 1/2`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit :
`lim_(x -> 7) [(x^3 - 343)/(sqrt(x) - sqrt(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 -> -3) (3x + 2)` = – 7
Evaluate the following :
`lim_(x -> 0)[x/(|x| + x^2)]`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)`
| x | – 3.1 | – 3.01 | – 3.00 | – 2.999 | – 2.99 | – 2.9 |
| f(x) | – 0.24845 | – 0.24984 | – 0.24998 | – 0.25001 | – 0.25015 | – 0.25158 |
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) (x^2 + 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 -> 3) 1/(x - 3)`
Verify the existence of `lim_(x -> 1) f(x)`, where `f(x) = {{:((|x - 1|)/(x - 1)",", "for" x ≠ 1),(0",", "for" x = 1):}`
Evaluate the following limits:
`lim_(x -> 2) (1/x - 1/2)/(x - 2)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 + x) - 1)/x`
Evaluate the following limits:
`lim_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1)`
Evaluate the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
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 -> 0) (sinalphax)/(sinbetax)`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
If `lim_(x -> 0) (sin "p"x)/(tan 3x)` = 4, then the value of p is
Choose the correct alternative:
`lim_(x -> 0) ("e"^(sin x) - 1)/x` =
