Advertisements
Advertisements
Question
Evaluate the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
Advertisements
Solution
`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
RELATED QUESTIONS
Evaluate the following limit :
`lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
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 :
`lim_(x -> 0)[x/(|x| + x^2)]`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 2) (x - 2)/(x^2 - 4)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.25641 | 0.25062 | 0.250062 | 0.24993 | 0.24937 | 0.24390 |
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)`
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 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_(x -> 1) (sqrt(x) - x^2)/(1 - sqrt(x))`
Evaluate the following limits:
`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) ("a" > "b")`
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 -> 0)(1 + x)^(1/(3x))`
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) (2 "arc"sinx)/(3x)`
Evaluate the following limits:
`lim_(x -> pi) (sin3x)/(sin2x)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 + sinx) - sqrt(1 - sinx))/tanx`
Choose the correct alternative:
`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` 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:
The value of `lim_(x -> 0) sinx/sqrt(x^2)` is
`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.
