Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x -> pi) (sin3x)/(sin2x)`
Advertisements
उत्तर
`lim_(x -> pi) (sin3x)/(sin2x) = lim_(x -> pi) (3sin x - 4 sin^3 x)/(2sinx cos x)`
= `lim_(x -> pi) [(3sinx)/(2sinx cosx) - (4sin^3x)/(sinx cosx)]`
= `lim_(x -> pi) [3/(2cosx) - (2sin^2x)/cosx]`
= `lim_(x -> pi) 3/(2cosx) - lim_(x -> pi) (2sin^2x)/cosx`
= `3/(2cospi) - (2sin^2pi)/cospi`
=`3/(2 xx -1) - (2 xx 0)/(-1)`
`lim_(x -> pi) (sin3x)/(sin2x) = - 3/2`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(z -> -3) [sqrt("z" + 6)/"z"]`
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
Evaluate the following :
`lim_(x -> 0) [(sqrt(1 - cosx))/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 -> 2) f(x)` where `f(x) = {{:(4 - x",", x ≠ 2),(0",", x = 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 -> 1) sin pi 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 -> x/2) tan x`
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_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + 1) - 1)/(sqrt(x^2 + 16) - 4)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 - x) - 1)/x^2`
Evaluate the following limits:
`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) ("a" > "b")`
Evaluate the following limits:
`lim_(x -> oo) 3/(x - 2) - (2x + 11)/(x^2 + x - 6)`
Evaluate the following limits:
`lim_(x -> oo) (x^3 + x)/(x^4 - 3x^2 + 1)`
Evaluate the following limits:
`lim_(x - oo){x[log(x + "a") - log(x)]}`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(2) - sqrt(1 + cosx))/(sin^2x)`
Evaluate the following limits:
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`
Choose the correct alternative:
`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =
`lim_(x -> 0) ((2 + x)^5 - 2)/((2 + x)^3 - 2)` = ______.
If f(x) is differentiable at x = 1 and `lim_(h → 0) 1/h f(1 + h) = 5`, then f' (1) is equal to ______.
