Advertisements
Advertisements
Question
Evaluate the following limits:
`lim_(x -> ) (sinx(1 - cosx))/x^3`
Solution
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
RELATED QUESTIONS
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) [(x^3 - 343)/(sqrt(x) - sqrt(7))]`
In problems 1 – 6, using the table estimate the value of the limit.
`lim_(x -> 2) (x - 2)/(x^2 - x - 2)`
x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
f(x) | 0.344820 | 0.33444 | 0.33344 | 0.333222 | 0.33222 | 0.332258 |
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`
x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
f(x) | 0.2911 | 0.2891 | 0.2886 | 0.2886 | 0.2885 | 0.28631 |
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) (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 -> 5) |x - 5|/(x - 5)`
Evaluate the following limits:
`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`
Evaluate the following limits:
`lim_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 - x) - 1)/x^2`
Evaluate the following limits:
`lim_(x ->oo) (x^3/(2x^2 - 1) - x^2/(2x + 1))`
Evaluate the following limits:
`lim_(x -> 0) (sinalphax)/(sinbetax)`
Evaluate the following limits:
`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(2) - sqrt(1 + cosx))/(sin^2x)`
Choose the correct alternative:
`lim_(x -> oo) sinx/x`
Choose the correct alternative:
`lim_(x -> 3) [x]` =
Choose the correct alternative:
`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is
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→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.