Advertisements
Advertisements
प्रश्न
Evaluate the following limit :
`lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
Advertisements
उत्तर
Put 1 – x = y
Then as x → 0, y → 1
∴ `lim_(x -> 0) ((1 - x)^8 - 1)/((1 - x)^2 - 1)`
= `lim_(y -> 1) (y^8 - 1)/(y^2 - 1)`
= `lim_(y -> 1)[(((y^8 - 1)/(y - 1)))/(((y^2 - 1)/(y - 1)))]` ...[∵ y → 1, y ≠ 1, ∴ y – 1 ≠ 0]
= `(lim_(y -> 1)((y^8 - 1)/(y - 1)))/(lim_(y -> 1)((y^2 - 1)/(y - 1))`
= `(8(1)^7)/(2(1)^1) ...[because lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`
= 4
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(x -> 3)[sqrt(2x + 6)/x]`
Evaluate the following limit:
`lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
Evaluate the following limit :
If `lim_(x -> 5) [(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
Evaluate the following :
Find the limit of the function, if it exists, at x = 1
f(x) = `{(7 - 4x, "for", x < 1),(x^2 + 2, "for", x ≥ 1):}`
Evaluate the following :
Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`
In problems 1 – 6, using the table estimate the value of the limit.
`lim_(x -> 2) (x - 2)/(x^2 - x - 2)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.344820 | 0.33444 | 0.33344 | 0.333222 | 0.33222 | 0.332258 |
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 |
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 -> 3) (4 - 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 -> 1) f(x)` where `f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", 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 -> 5) |x - 5|/(x - 5)`
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`
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 -> 0) sec 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`
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):}`
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.
If the limit of f(x) as x approaches 2 is 4, can you conclude anything about f(2)? Explain reasoning
Verify the existence of `lim_(x -> 1) f(x)`, where `f(x) = {{:((|x - 1|)/(x - 1)",", "for" x ≠ 1),(0",", "for" x = 1):}`
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 - 0) (sqrt(1 + x^2) - 1)/x`
Evaluate the following limits:
`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) ("a" > "b")`
Find the left and right limits of f(x) = tan x at x = `pi/2`
Evaluate the following limits:
`lim_(x -> oo) (x^4 - 5x)/(x^2 - 3x + 1)`
Show that `lim_("n" -> oo) (1^2 + 2^2 + ... + (3"n")^2)/((1 + 2 + ... + 5"n")(2"n" + 3)) = 9/25`
Evaluate the following limits:
`lim_(x -> oo)(1 + 1/x)^(7x)`
Evaluate the following limits:
`lim_(x -> 0)(1 + x)^(1/(3x))`
Evaluate the following limits:
`lim_(x -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`
Evaluate the following limits:
`lim_(x -> 0) (sinalphax)/(sinbetax)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 + sinx) - sqrt(1 - sinx))/tanx`
Evaluate the following limits:
`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 -4x + 2))^x`
Evaluate the following limits:
`lim_(x -> 0) (tan x - sin x)/x^3`
Choose the correct alternative:
`lim_(theta -> 0) (sinsqrt(theta))/(sqrt(sin theta)`
Choose the correct alternative:
`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =
Choose the correct alternative:
`lim_(x -> 0) ("a"^x - "b"^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 -> 0) (sin 4x + sin 2x)/(sin5x - sin3x)` = ______.
