Advertisements
Advertisements
प्रश्न
Evaluate the following limit:
`lim_(y -> -3) [(y^5 + 243)/(y^3 + 27)]`
Advertisements
उत्तर
`lim_(y -> -3) [(y^5 + 243)/(y^3 + 27)]`
= `lim_(y -> -3) (((y^5 + 243)/(y + 3)))/(((y^3 + 27)/(y + 3))) ...[(because y -> -3"," therefore y ≠ -3","),(therefore y + 3 ≠ 0)]`
= `(lim_(y -> -3)[(y^5 - (- 3)^5)/(y - ( - 3))])/(lim_(y -> - 3)[(y^3 - ( - 3)^3)/(y - (- 3))]`
= `(5(-3)^4)/(3(-3)^2) ...[because lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "n"*"a"^("n" - 1)]`
= `5/3 xx 9`
= 15
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`
Evaluate the following limit:
If `lim_(x -> 1)[(x^4 - 1)/(x - 1)]` = `lim_(x -> "a")[(x^3 - "a"^3)/(x - "a")]`, find all possible values of a
Evaluate the following limit :
`lim_(x -> 1)[(x + x^2 + x^3 + ......... + x^"n" - "n")/(x - 1)]`
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 -> 2)(2x + 3)` = 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 -> -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 -> 2) (x^2 - 1)` = 3
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 -> - 3) (sqrt(1 - x) - 2)/(x + 3)`
| x | – 3.1 | – 3.01 | – 3.00 | – 2.999 | – 2.99 | – 2.9 |
| f(x) | – 0.24845 | – 0.24984 | – 0.24998 | – 0.25001 | – 0.25015 | – 0.25158 |
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 -> 3) 1/(x - 3)`
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)`
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):}`
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
Evaluate the following limits:
`lim_(x ->) (x^"m" - 1)/(x^"n" - 1)`, m and n are integers
Evaluate the following limits:
`lim_(x -> 1) (sqrt(x) - x^2)/(1 - sqrt(x))`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + 1) - 1)/(sqrt(x^2 + 16) - 4)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 - x) - 1)/x^2`
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)`
Evaluate the following limits:
`lim_(x -> 0) (sinalphax)/(sinbetax)`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/x`
Evaluate the following limits:
`lim_(x -> 0) (2^x - 3^x)/x`
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 -> ) (sinx(1 - cosx))/x^3`
Choose the correct alternative:
`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =
Choose the correct alternative:
`lim_(x -> 0) (8^x - 4x - 2^x + 1^x)/x^2` =
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 -> oo) (1/"n"^2 + 2/"n"^2 + 3/"n"^2 + ... + "n"/"n"^2)` is
Choose the correct alternative:
`lim_(x -> 0) ("e"^(sin x) - 1)/x` =
If `lim_(x -> 1) (x + x^2 + x^3|+ .... + x^n - n)/(x - 1)` = 820, (n ∈ N) then the value of n is equal to ______.
`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.
