Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 -4x + 2))^x`
Advertisements
उत्तर
`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 -4x + 2))^x = lim_(x -> oo) ((x^2 - 4x + 2 + 2x - 1)/(x^2 - 4x + 2))^x`
= `lim_(x -> oo) [(x^2 - 4x - 2)/(x^2 - 4x + 2) +(2x - 1)/(x^2 - 4x + 2)]^x`
= `lim_(x -> oo) [1 + (2x - 1)/(x^2 - 4x + 2)]^x`
= `lim_(x - oo) [1 + 1/((x^2 -4x + 2)/(2x - 1))]^(((x^2 - 4x + 2)/(2x - 1) xx ((2x - 1)x)/(x^2 - 4x + 2))`
= `lim_(x -> oo) [(1 + (2x - 1)/(x^2 - 4x + 2))^((x^2 - 4x + 2)/(2x - 1))]^(((2x - 1)x)/(x^2 - 4x + 2))`
`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 - 4x + 2))^x = [lim_(x -> oo) "e"]^((2x^2 - x)/(x^2 - 4x + 2))`
`lim_(x -> oo) (1 + 1/x)^x` = e
= `"e"^(lim_(x ->oo)) ((2x^2 - x)/(x^2 - 4x + 2))`
= `"e"^(lim_(x -> oo) (x^2(2 -x/x^2))/(x^2(1 - (4x)/(x^2) + 2/x^2))`
= `"e"^(lim_(x ->oo) ((2 - 1/x)/(1 -4/x + 2/x^2))`
= `"e"^(((2 - 0)/(1 - 0 + 0))`
`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 -4x + 2))^x` = e2
APPEARS IN
संबंधित प्रश्न
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) (x^2 - 1)` = 3
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
Evaluate the following :
`lim_(x -> 0) {1/x^12 [1 - cos(x^2/2) - cos(x^4/4) + cos(x^2/2) cos(x^4/4)]}`
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 problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (cos x - 1)/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.0001 | 0.01 | 0.1 |
| f(x) | 0.04995 | 0.0049999 | 0.0004999 | – 0.0004999 | – 0.004999 | – 0.04995 |
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):}`
Evaluate the following limits:
`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`
Evaluate the following limits:
`lim_(x -> oo) (x^3 + x)/(x^4 - 3x^2 + 1)`
Evaluate the following limits:
`lim_(x -> pi) (sin3x)/(sin2x)`
Evaluate the following limits:
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`
Evaluate the following limits:
`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =
If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.
`lim_(x -> 5) |x - 5|/(x - 5)` = ______.
`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.
The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is ______.
The value of `lim_(x rightarrow 0) (sqrt((1 + x^2)) - sqrt(1 - x^2))/x^2` is ______.
