हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

Evaluate the following limits: limx→∞(1+1x)7x

Advertisements
Advertisements

प्रश्न

Evaluate the following limits:

`lim_(x -> oo)(1 + 1/x)^(7x)`

योग
Advertisements

उत्तर

We know `lim_(x -> oo)(x + 1/x)^x` = e

`lim_(x -> oo)(1 + 1/x)^(7x) = [lim_(x -> oo)(1 + 1/x)^x]^7`

= e7 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.4 [पृष्ठ ११७]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.4 | Q 1 | पृष्ठ ११७

संबंधित प्रश्न

Evaluate the following limit :

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


Evaluate the following limit :

`lim_(y -> 1)[(2y - 2)/(root(3)(7 + y) - 2)]`


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 :

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):}`


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) (x^2 + 2)`


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

f(0) is undefined

`lim_(x -> 0) f(x)` = 4

f(2) = 6

`lim_(x -> 2) f(x)` = 3


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 -> oo) (x^4 - 5x)/(x^2 - 3x + 1)`


Show that `lim_("n" -> oo) (1 + 2 + 3 + ... + "n")/(3"n"^2 + 7n" + 2) = 1/6`


Evaluate the following limits:

`lim_(x -> 0) (sinalphax)/(sinbetax)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> ) (sinx(1 - cosx))/x^3`


Choose the correct alternative:

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


Choose the correct alternative:

`lim_(x -> 0) (8^x - 4x - 2^x + 1^x)/x^2` =


Choose the correct alternative:

`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×