English

Evaluate the following limit: limx→3[2x+6x] - Mathematics and Statistics

Advertisements
Advertisements

Question

Evaluate the following limit:

`lim_(x -> 3)[sqrt(2x + 6)/x]`

Sum

Solution

`lim_(x -> 3)[sqrt(2x + 6)/x]`

= `(lim_(x -> 3) sqrt(2x + 6))/(lim_(x -> 3) x`

= `sqrt(2(3) + 6)/3`

= `sqrt(12)/3`

= `(2sqrt(3))/3`

shaalaa.com
Concept of Limits
  Is there an error in this question or solution?
Chapter 7: Limits - Exercise 7.1 [Page 138]

RELATED QUESTIONS

Evaluate the following limit:

`lim_(z -> -3) [sqrt("z" + 6)/"z"]`


Evaluate the following limit:

`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`


Evaluate the following limit :

`lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`


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)(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


Evaluate the following :

`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(x - 1)]`


Evaluate the following :

`lim_(x -> 0) {1/x^12 [1 - cos(x^2/2) - cos(x^4/4) + cos(x^2/2) cos(x^4/4)]}`


In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) sin x/x`

x – 0.1  – 0.01 – 0.001 0.001 0.01 0.1
f(x) 0.99833 0.99998 0.99999 0.99999 0.99998 0.99833

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):}`


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)`


Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.

f(x) = `{{:(x^2",", x ≤ 2),(8 - 2x",", 2 < x < 4),(4",", x ≥ 4):}`


Evaluate the following limits:

`lim_(x ->) (x^"m" - 1)/(x^"n" - 1)`, m and n are integers


Evaluate the following limits:

`lim_(sqrt(x) -> 3) (x^2 - 81)/(sqrt(x) - 3)`


Evaluate the following limits:

`lim_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1)`


Evaluate the following limits:

`lim_(x - 0) (sqrt(1 + x^2) - 1)/x`


Evaluate the following limits:

`lim_(x -> 0) (sqrt(1 - x) - 1)/x^2`


Evaluate the following limits:

`lim_(x -> oo) (1 + x - 3x^3)/(1 + x^2 +3x^3)`


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 -> oo)(1 + "k"/x)^("m"/x)`


Evaluate the following limits:

`lim_(x -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`


Evaluate the following limits:

`lim_(x -> 0) (tan 2x)/(sin 5x)`


Evaluate the following limits:

`lim_(x -> 0) (sin("a" + x) - sin("a" - x))/x`


Evaluate the following limits:

`lim_(x -> 0) (2 "arc"sinx)/(3x)`


Evaluate the following limits:

`lim_(x -> 0) (2^x - 3^x)/x`


Evaluate the following limits:

`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`


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 -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is


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


If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.


`lim_(x -> 0) ((2 + x)^5 - 2)/((2 + x)^3 - 2)` = ______.


`lim_(x -> 5) |x - 5|/(x - 5)` = ______.


`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 ______.


`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×