Advertisements
Advertisements
Question
Evaluate the following limits:
`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`
Advertisements
Solution
`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`
Pu y = x + 4
⇒ x = y – 4
⇒ x – 5 = y – 4 – 5
⇒ x – 5 = y – 9
⇒ y → 5 + 4 = 9
∴ `lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5) = lim_(y -> 9) (sqrt(y) - sqrt(3^2))/(y - 9)`
= `lim_(y -> 9) (y^(1/2) - (9)^(1/2))/(y - 9)`
`lim_(x -> "a") (x^"n" - "a"^"n") = "na"^("n" - 1)`
`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5) = 1/2(9)^(1/2 - 1)`
= `1/2 (9)^(-1/2)`
= `1/2 xx 1/(9^(1/2)`
= `1/2 xx 1/sqrt(9)`
= `1/2 xx 1/3`
= `1/6`
APPEARS IN
RELATED QUESTIONS
Evaluate the following limit:
`lim_(y -> -3) [(y^5 + 243)/(y^3 + 27)]`
Evaluate the following limit :
`lim_(x -> 1)[(x + x^2 + x^3 + ......... + x^"n" - "n")/(x - 1)]`
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 -> 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 -> 1) f(x)` where `f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", x = 1):}`
Verify the existence of `lim_(x -> 1) f(x)`, where `f(x) = {{:((|x - 1|)/(x - 1)",", "for" x ≠ 1),(0",", "for" x = 1):}`
Find the left and right limits of f(x) = tan x at x = `pi/2`
Evaluate the following limits:
`lim_(x -> oo) 3/(x - 2) - (2x + 11)/(x^2 + x - 6)`
Evaluate the following limits:
`lim_(x -> oo) (x^3 + x)/(x^4 - 3x^2 + 1)`
Evaluate the following limits:
`lim_(x -> oo) (1 + 3/x)^(x + 2)`
Evaluate the following limits:
`lim_(x -> 0) (2 "arc"sinx)/(3x)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 + sinx) - sqrt(1 - sinx))/tanx`
Choose the correct alternative:
`lim_(x -> oo) sinx/x`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =
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 ______.
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 ______.
The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is ______.
