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:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following limit :
If `lim_(x -> 5) [(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
Evaluate the following limit :
`lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
Evaluate the following limit :
`lim_(y -> 1)[(2y - 2)/(root(3)(7 + y) - 2)]`
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
Evaluate the following :
`lim_(x -> 0) [(sqrt(1 - cosx))/x]`
Evaluate the following :
`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(x - 1)]`
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 |
Evaluate the following limits:
`lim_(x ->) (x^"m" - 1)/(x^"n" - 1)`, m and n are integers
Evaluate the following limits:
`lim_(x -> 2) (1/x - 1/2)/(x - 2)`
Evaluate the following limits:
`lim_(x -> 1) (sqrt(x) - x^2)/(1 - sqrt(x))`
Evaluate the following limits:
`lim_(x -> oo) (1 + 3/x)^(x + 2)`
Evaluate the following limits:
`lim_(x-> 0) (1 - cos x)/x^2`
Evaluate the following limits:
`lim_(x -> 0) (2^x - 3^x)/x`
Evaluate the following limits:
`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`
Choose the correct alternative:
`lim_(x -> 3) [x]` =
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
Choose the correct alternative:
If `lim_(x -> 0) (sin "p"x)/(tan 3x)` = 4, then the value of p is
`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.
`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.
