English
Tamil Nadu Board of Secondary EducationHSC Science Class 11

Choose the correct alternative: limx→08x-4x-2x+1xx2 =

Advertisements
Advertisements

Question

Choose the correct alternative:

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

Options

  • 2 log 2

  • 2(log)2 

  • log 2

  • 3 log 2

MCQ
Advertisements

Solution

2(log)2 

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Differential Calculus - Limits and Continuity - Exercise 9.6 [Page 130]

APPEARS IN

Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 11 TN Board
Chapter 9 Differential Calculus - Limits and Continuity
Exercise 9.6 | Q 8 | Page 130

RELATED QUESTIONS

Evaluate the following limit:

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


Evaluate the following limit :

`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`


Evaluate the following :

`lim_(x -> 0)[x/(|x| + x^2)]`


Evaluate the following :

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


In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)`

x – 3.1  – 3.01 – 3.00 – 2.999 – 2.99 – 2.9
f(x) – 0.24845 – 0.24984 – 0.24998 – 0.25001 – 0.25015 – 0.25158

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):}`


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.


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 -> 2) (x^4 - 16)/(x - 2)`


Evaluate the following limits:

`lim_(x -> 0) (sqrt(1 + x) - 1)/x`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9))`


Evaluate the following limits:

`lim_(x  -> oo) 3/(x - 2) - (2x + 11)/(x^2 + x - 6)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Choose the correct alternative:

`lim_(x -> 0) ("e"^(sin x) - 1)/x` =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×