Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x - oo){x[log(x + "a") - log(x)]}`
Advertisements
उत्तर
We Know `lim_(x -> 0) (log(1 +x))/x` = 1
`lim_(x - oo){x[log(x + "a") - log(x)]}`
= `lim_(x > oo) x*log((x + "a")/x)`
= `lim_(x -> oo) log(x/x + "a"/x)/(1/x)`
= `lim_(x -> oo) (log(1 + "a"/x))/(1/"a" xx "a"/x)`
= `"a" lim_(x -> oo) (log (1 + "a"/x))/("a"/x)` ......(1)
Put y = `"a"/x`
When x = `oo` then y = `"a"/oo` = 0
x → `oo`
⇒ y → 0
(1) ⇒ `lim_(x - oo){x[log(x + "a") - log(x)]}`
= `"a" lim_(y -> 0) (log(1 + y))/y`
= a × 1
= a
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit :
If `lim_(x -> 5) [(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> -3) (3x + 2)` = – 7
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) = `{{:(x^2",", x ≤ 2),(8 - 2x",", 2 < x < 4),(4",", x ≥ 4):}`
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Evaluate : `lim_(x -> 3) (x^2 - 9)/(x - 3)` if it exists by finding `f(3^-)` and `f(3^+)`
Evaluate the following limits:
`lim_(sqrt(x) -> 3) (x^2 - 81)/(sqrt(x) - 3)`
Evaluate the following limits:
`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`
Evaluate the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
Evaluate the following limits:
`lim_(x -> oo) (1 + x - 3x^3)/(1 + x^2 +3x^3)`
Evaluate the following limits:
`lim_(x -> 0)(1 + x)^(1/(3x))`
Evaluate the following limits:
`lim_(x -> 0) (sin^3(x/2))/x^2`
Evaluate the following limits:
`lim_(x -> 0) (sin("a" + x) - sin("a" - x))/x`
Evaluate the following limits:
`lim_(x -> 0) (2 "arc"sinx)/(3x)`
Evaluate the following limits:
`lim_(x-> 0) (1 - cos x)/x^2`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/x`
Choose the correct alternative:
`lim_(x -> 0) (8^x - 4x - 2^x + 1^x)/x^2` =
Choose the correct alternative:
`lim_(x -> 0) ("e"^tanx - "e"^x)/(tan x - x)` =
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)` = ______.
