Advertisements
Advertisements
Question
Evaluate the following limits:
`lim_(x -> 0) (sinalphax)/(sinbetax)`
Advertisements
Solution
We know `lim_(x -> 0) sinx/x` = 1
`lim_(x -> 0) (sin alpha x)/(sin betax) = lim_(x -> 0) (sin alphax)/(1/alpha (alphax)) xx (1/beta (betax))/(sin betax)`
= `alpha/beta lim_(x -> 0) (sin(alphax))/((alphax)) xx (betax)/(sin(betax))`
= `alpha/beta lim_(alphax -> 0) (sin(alphax))/(alphax) xx lim_(betax -> 0) (betax)/(sin(betax))`
= `alpha/beta lim_(alphax -> 0) (sin(alphax))/(alphax) xx 1/(lim_(betax -> 0) (sin("betax))/(betax))`
= `alpha/beta xx 1 xx 1/1`
`lim_(x -> 0) (sinalphax)/(sinbetax) = alpha/beta`
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 -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
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 -> 1) (x^2 + x + 1)` = 3
Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.
f(x) = `{{:(sin x",", x < 0),(1 - cos x",", 0 ≤ x ≤ pi),(cos x",", x > pi):}`
Write a brief description of the meaning of the notation `lim_(x -> 8) f(x)` = 25
If the limit of f(x) as x approaches 2 is 4, can you conclude anything about f(2)? Explain reasoning
Evaluate the following limits:
`lim_(x - 0) (sqrt(1 + x^2) - 1)/x`
Evaluate the following limits:
`lim_(x -> oo) (x^3 + x)/(x^4 - 3x^2 + 1)`
Evaluate the following limits:
`lim_(x -> oo) (1 + 3/x)^(x + 2)`
Evaluate the following limits:
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`
Evaluate the following limits:
`lim_(x -> 0) (2 "arc"sinx)/(3x)`
Evaluate the following limits:
`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`
Evaluate the following limits:
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`
Choose the correct alternative:
`lim_(x -> 0) ("a"^x - "b"^x)/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 -> oo) (1/"n"^2 + 2/"n"^2 + 3/"n"^2 + ... + "n"/"n"^2)` is
The value of `lim_(x rightarrow 0) (sqrt((1 + x^2)) - sqrt(1 - x^2))/x^2` is ______.
