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Choose the correct alternative: limx→01-cos2xx - Mathematics

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प्रश्न

Choose the correct alternative:

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

पर्याय

  • 0

  • 1

  • `sqrt(2)`

  • does not exist

MCQ
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उत्तर

`sqrt(2)`

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Differential Calculus - Limits and Continuity - Exercise 9.6 [पृष्ठ १२९]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.6 | Q 3 | पृष्ठ १२९

संबंधित प्रश्‍न

Evaluate the following limit :

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


Evaluate the following limit :

`lim_(x -> 7) [(x^3 - 343)/(sqrt(x) - sqrt(7))]`


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

x 1.9 1.99 1.999 2.001 2.01 2.1
f(x) 0.25641 0.25062 0.250062 0.24993 0.24937 0.24390

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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) ("a" > "b")`


Find the left and right limits of f(x) = `(x^2 - 4)/((x^2 + 4x+ 4)(x + 3))` at x = – 2


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


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) (2^x - 3^x)/x`


Evaluate the following limits:

`lim_(x -> pi) (sin3x)/(sin2x)`


Evaluate the following limits:

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


Choose the correct alternative:

`lim_(x - pi/2) (2x - pi)/cos x`


Choose the correct alternative:

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


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


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