Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
Evaluate the following limit:
`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`
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 -> 1) [(x + x^3 + x^5 + ... + x^(2"n" - 1) - "n")/(x - 1)]`
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) f(x)` where `f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", x = 1):}`
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.
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_("h" -> 0) (sqrt(x + "h") - sqrt(x))/"h", x > 0`
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 -> oo) 3/(x - 2) - (2x + 11)/(x^2 + x - 6)`
Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1
Evaluate the following limits:
`lim_(x -> 0) (sin^3(x/2))/x^2`
Evaluate the following limits:
`lim_(x -> 0) (2 "arc"sinx)/(3x)`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/x`
Evaluate the following limits:
`lim_(x -> pi) (sin3x)/(sin2x)`
Choose the correct alternative:
`lim_(x -> 0) ("e"^(sin x) - 1)/x` =
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)` = ______.
