Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 - x) - 1)/x^2`
Advertisements
उत्तर
`lim_(x -> 0) (sqrt(1 - x) - 1)/x^2 = lim_(x -> 0) [(sqrt(1 - x) - 1)/x^2 xx (sqrt(1 - x) + 1)/(sqrt(1 - x) + 1)]`
= `lim_( -> 0) [((1 - x) - 1)/(x^2 (sqrt(1 - x) + 1))]`
= `lim_(x -> 0) [(- x)/(x(sqrt(1 - x) + 1))]`
= `- lim_(x -> 0) [1/(x(sqrt(1 - x) + 1))]`
= `- 1/(0(sqrt(1 - 0) + 1))`
= `- oo`
∴ `lim_(x -> 0) (sqrt(1 - x) - 1)/x^2` does not exist.
APPEARS IN
संबंधित प्रश्न
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 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 :
Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`
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 -> x/2) tan x`
If the limit of f(x) as x approaches 2 is 4, can you conclude anything about f(2)? Explain reasoning
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) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`
Evaluate the following limits:
`lim_(x-> 0) (1 - cos x)/x^2`
Evaluate the following limits:
`lim_(x -> pi) (1 + sinx)^(2"cosec"x)`
Choose the correct alternative:
`lim_(x - pi/2) (2x - pi)/cos x`
Choose the correct alternative:
`lim_(x -> 0) sqrt(1 - cos 2x)/x`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
If `f(x) = x(- 1)^([1/x])`, x ≤ 0, then the value of `lim_(x -> 0) f(x)` is equal to
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
Choose the correct alternative:
`lim_(x -> 0) ("e"^(sin x) - 1)/x` =
`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.
If f(x) is differentiable at x = 1 and `lim_(h → 0) 1/h f(1 + h) = 5`, then f' (1) is equal to ______.
