English

Evaluate the following limit: limz→-5[(1z+15)z+5] - Mathematics and Statistics

Advertisements
Advertisements

Question

Evaluate the following limit:

`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`

Sum
Advertisements

Solution

`lim_(z -> -5)[(1/z + 1/5)/(z + 5)]`

= `lim_(z -> -5) (((5 + z)/(5z)))/(z + 5)`

= `lim_(z -> - 5) 1/(5z)`    ...[∵ z → – 5, z ≠ – 5 ∴ z + 5 ≠ 0]

= `(lim_(z -> - 5)(1))/(lim_(z -> - 5) (5z))`

= `1/(5( - 5))`

= `-1/25`

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

RELATED QUESTIONS

Evaluate the following limit:

`lim_(y -> -3) [(y^5 + 243)/(y^3 + 27)]`


Evaluate the following limit:

`lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`


Evaluate the following limit:

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


Evaluate the following limit :

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


Evaluate the following limit : 

If `lim_(x -> 5) [(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.


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 :

Find the limit of the function, if it exists, at x = 1

f(x) = `{(7 - 4x, "for", x < 1),(x^2 + 2, "for", 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) (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 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 -> 3) (4 - x)`


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):}`


Sketch the graph of a function f that satisfies the given value:

f(– 2) = 0

f(2) = 0

`lim_(x -> 2) f(x)` = 0

`lim_(x -> 2) f(x)` does not exist.


Evaluate the following limits:

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


Evaluate the following limits:

`lim_("h" -> 0) (sqrt(x + "h") - sqrt(x))/"h", x > 0`


Evaluate the following limits:

`lim_(x -> 0) (sqrt(x^2 + 1) - 1)/(sqrt(x^2 + 16) - 4)`


Evaluate the following limits:

`lim_(x -> 5) (sqrt(x - 1) - 2)/(x - 5)`


Evaluate the following limits:

`lim_(x -> oo) (x^3 + x)/(x^4 - 3x^2 + 1)`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0)(1 + x)^(1/(3x))`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x - oo){x[log(x + "a") - log(x)]}`


Evaluate the following limits:

`lim_(x -> pi) (1 + sinx)^(2"cosec"x)`


Evaluate the following limits:

`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`


Choose the correct alternative:

`lim_(x -> oo) sinx/x`


Choose the correct alternative:

`lim_(x -> 0) ("a"^x - "b"^x)/x` =


Choose the correct alternative:

`lim_(x -> oo) (1/"n"^2 + 2/"n"^2 + 3/"n"^2 + ... + "n"/"n"^2)` is


Choose the correct alternative:

`lim_(x -> 0) ("e"^tanx - "e"^x)/(tan x - x)` =


`lim_(x -> 0) (sin 4x + sin 2x)/(sin5x - sin3x)` = ______.


The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×