Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
Advertisements
उत्तर
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x)) = lim_(x -> 2) (sqrt(x + 2) - 2)/(root(3)(4 - x) - root(3)(2))`
= `lim_(x -> 2) ((x + 2)^(1/2) - (2^2)^(1/2))/((4 - x)^(1/3) - (2)^(1/3))`
= `lim_(x -> 2) ((x + 2)^(1/2) - (2)^(1/2))/(x - 2) xx (x - 2)/((4 - x)^(1/3) - (2)^(1/3))`
= `lim_(x -> 2) ((x + 2)^(1/2) - (4)^(1/2))/((x + 2) - 4) xx (-[(4 - x) - 2])/((4 - x)^(1/3) - (2)^(1/3)]`
= `lim_(x -> 2) ((x + 2)^(1/2)- (4)^(1/2))/((x + 2) - 4) xx - 1/(lim_(x -> 2) ((4 - x)^(1/3) - (2)^(1/3))/((4 - x) - 2)`
`lim_(x -> "a") (x^"n" - "a"^"n") = "na"^("n" - 1)`
= `1/2(4)^(1/2 - 1) xx - 1/(1/3 (2)^(1/3 - 1)`
= `1/2(4)^(-1/2) xx - 3/((2)^(-2/3)`
= `1/(2(2^2)^(1/2)) xx - 3 xx 2^(2/3)`
= `- 1/(2 xx 2) xx 3 xx 2^(2/3)`
= ` - 3/4 xx (2^2)^(1/3)`
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x)) = - 3/4 root(3)(4)`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(x -> 3)[sqrt(2x + 6)/x]`
Evaluate the following limit :
`lim_(y -> 1)[(2y - 2)/(root(3)(7 + y) - 2)]`
Evaluate the following limit :
`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) | 0.2911 | 0.2891 | 0.2886 | 0.2886 | 0.2885 | 0.28631 |
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 -> 2) f(x)` where `f(x) = {{:(4 - x",", x ≠ 2),(0",", x = 2):}`
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_("h" -> 0) (sqrt(x + "h") - sqrt(x))/"h", x > 0`
Evaluate the following limits:
`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`
Evaluate the following limits:
`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9))`
Evaluate the following limits:
`lim_(x -> oo) 3/(x - 2) - (2x + 11)/(x^2 + x - 6)`
Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1
Evaluate the following limits:
`lim_(x -> 0)(1 + x)^(1/(3x))`
Evaluate the following limits:
`lim_(x -> oo) (1 + 3/x)^(x + 2)`
Evaluate the following limits:
`lim_(x -> 0) (sinalphax)/(sinbetax)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + "a"^2) - "a")/(sqrt(x^2 + "b"^2) - "b")`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(2) - sqrt(1 + cosx))/(sin^2x)`
Evaluate the following limits:
`lim_(x -> ) (sinx(1 - cosx))/x^3`
Choose the correct alternative:
`lim_(x -> 3) [x]` =
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.
