Advertisements
Advertisements
Question
Evaluate the following limits:
`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) ("a" > "b")`
Advertisements
Solution
`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2), "a" > "b"`
`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) = lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) xx (sqrt(x - "b") + sqrt("a" - "b"))/(sqrt(x - "b") + sqrt("a" - "b"))`
= `lim_(x -> "a") ((x - "b") - ("a"- "b"))/((x^2 - "a"^2) [sqrt(x - "b") + sqrt("a" - "b")]`
= `lim_(x -> "a") (x - "b" - "a" + "b")/((x - "a")(x + "a") [sqrt(x - "b") + sqrt("a" - "b")]`
= `lim_(x -> "a") (x - "a")/((x - "a")(x + "a") [sqrt(x - "b") + sqrt("a" - "b")]`
= `lim_(x -> "a") 1/((x + "a")[sqrt("x" - "b") + sqrt('a" -"b")]`
= `1/(("a" + "a")[sqrt("a" - "b") + sqrt("a" - "b")]`
= `1/(2"a" xx 2sqrt("a" - "b")`
= `1/(4"a"sqrt("a" - "b")`
APPEARS IN
RELATED QUESTIONS
Evaluate the following limit:
`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`
Evaluate the following limit:
`lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
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 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
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 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 -> 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)`
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 ->) (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))`
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 + "k"/x)^("m"/x)`
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` =
The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is ______.
