Advertisements
Advertisements
Question
Choose the correct alternative:
If `lim_(x -> 0) (sin "p"x)/(tan 3x)` = 4, then the value of p is
Options
6
9
12
4
Solution
12
APPEARS IN
RELATED QUESTIONS
Evaluate the following limit:
`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`
Evaluate the following limit :
`lim_(y -> 1)[(2y - 2)/(root(3)(7 + y) - 2)]`
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 -> 2)(2x + 3)` = 7
Sketch the graph of a function f that satisfies the given value:
f(0) is undefined
`lim_(x -> 0) f(x)` = 4
f(2) = 6
`lim_(x -> 2) f(x)` = 3
Evaluate the following limits:
`lim_(x ->oo) (x^3/(2x^2 - 1) - x^2/(2x + 1))`
Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1
An important problem in fishery science is to estimate the number of fish presently spawning in streams and use this information to predict the number of mature fish or “recruits” that will return to the rivers during the reproductive period. If S is the number of spawners and R the number of recruits, “Beverton-Holt spawner recruit function” is R(S) = `"S"/((alpha"S" + beta)` where `alpha` and `beta` are positive constants. Show that this function predicts approximately constant recruitment when the number of spawners is sufficiently large
Evaluate the following limits:
`lim_(x -> oo)(1 + "k"/x)^("m"/x)`
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 -> 0) (3^x - 1)/(sqrt(x + 1) - 1)`
Evaluate the following limits:
`lim_(x -> oo) x [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)]`
Evaluate the following limits:
`lim_(x - oo){x[log(x + "a") - log(x)]}`
Evaluate the following limits:
`lim_(x -> ) (sinx(1 - cosx))/x^3`
Choose the correct alternative:
If `f(x) = x(- 1)^([1/x])`, x ≤ 0, then the value of `lim_(x -> 0) f(x)` is equal to
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.
`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.