Advertisements
Advertisements
प्रश्न
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 |
Advertisements
उत्तर
Let f(x) = `sin x/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) |
`(sin(- 0.1))/(- 0.1)` = `(- sin(0.1))/(- 0.1)` = 0.998 |
`(sin(- 0.01))/(- 0.01)` = `(- sin(0.01))/(- 0.01)` = 0.999 |
`(sin(- 0.001))/(- 0.001)` = `(- sin(0.001))/(- 0.001)` = 0.9999 |
`(sin(0.001))/(0.001)` = 0.9999 |
`(sin(0.01))/(0.01)` = 0.999 |
`(sin(0.1))/(0.1)` = 0.998 |
`lim_(x -> 0) sin x/x` = 1
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit :
`lim_(x -> 1)[(x + x^2 + x^3 + ......... + x^"n" - "n")/(x - 1)]`
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 -> -3) (3x + 2)` = – 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
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 -> 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 -> 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 -> 3) 1/(x - 3)`
Evaluate : `lim_(x -> 3) (x^2 - 9)/(x - 3)` if it exists by finding `f(3^-)` and `f(3^+)`
Evaluate the following limits:
`lim_("h" -> 0) (sqrt(x + "h") - sqrt(x))/"h", x > 0`
Evaluate the following limits:
`lim_(x -> oo) (x^4 - 5x)/(x^2 - 3x + 1)`
Evaluate the following limits:
`lim_(x ->oo) (x^3/(2x^2 - 1) - x^2/(2x + 1))`
Evaluate the following limits:
`lim_(x -> oo)(1 + 1/x)^(7x)`
Evaluate the following limits:
`lim_(x -> 0) (sinalphax)/(sinbetax)`
Evaluate the following limits:
`lim_(x -> 0) (2^x - 3^x)/x`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
`lim_(x -> 3) [x]` =
Choose the correct alternative:
`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is
