Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`
Advertisements
उत्तर
We know `lim_(x -> 0) ("e"^x - 1)/x` = 1
`lim_(x -> 0) sinx/x` = 1
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx = lim_(x -> 0) ("e"^x - 1/"e"^x)/sinx`
= `lim_(x -> 0) (("e"^x * "e"^x - 1)/"e"^x)/(sinx)`
= `lim_(x -> 0) ("e"^(2x) - 1)/("e"^x sinx)`
= `lim_(x -> 0) (1/"e"^x xx ("e"^(2x) - 1)/(1/2 xx 2x) xx x/sinx)`
= `(lim_(x -> 0) 1/"e"^x) 2(lim_(2x -> 0) ("e"^(2x) - 1)/(2x)) xx 1/((lim_(x -> 0) sinx/x))`
= `1/"e"^0 xx 2 xx 1 xx 1/1`
= `1/1 xx 2 xx 1`
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx` = 2
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following :
Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`
Evaluate the following :
`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(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) sin pi x`
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
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.
Evaluate the following limits:
`lim_("h" -> 0) (sqrt(x + "h") - sqrt(x))/"h", x > 0`
Evaluate the following limits:
`lim_(x - 0) (sqrt(1 + x^2) - 1)/x`
Find the left and right limits of f(x) = tan x at x = `pi/2`
Evaluate the following limits:
`lim_(x -> oo)(1 + 1/x)^(7x)`
Evaluate the following limits:
`lim_(x -> oo)(1 + "k"/x)^("m"/x)`
Evaluate the following limits:
`lim_(x -> oo) (1 + 3/x)^(x + 2)`
Evaluate the following limits:
`lim_(x -> 0) (sin("a" + x) - sin("a" - x))/x`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/x`
Evaluate the following limits:
`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`
Evaluate the following limits:
`lim_(x -> 0) (tan x - sin x)/x^3`
Choose the correct alternative:
`lim_(x -> 0) sqrt(1 - cos 2x)/x`
Choose the correct alternative:
`lim_(x -> 0) ("e"^(sin x) - 1)/x` =
`lim_(x→∞)((x + 7)/(x + 2))^(x + 4)` is ______.
The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is ______.
