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Choose the correct alternative: If plimx→0sinpxtan3x = 4, then the value of p is - Mathematics

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प्रश्न

Choose the correct alternative:

If `lim_(x -> 0) (sin "p"x)/(tan 3x)` = 4, then the value of p is

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उत्तर

12

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.6 [पृष्ठ १३०]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.6 | Q 14 | पृष्ठ १३०

संबंधित प्रश्न

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 - x - 2)`

x 1.9 1.99 1.999 2.001 2.01 2.1
f(x) 0.344820 0.33444 0.33344 0.333222 0.33222 0.332258

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

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)`


Show that  `lim_("n" -> oo) (1^2 + 2^2 + ... + (3"n")^2)/((1 + 2 + ... + 5"n")(2"n" + 3)) = 9/25`


Evaluate the following limits:

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


Evaluate the following limits:

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


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) (3^x - 1)/(sqrt(x + 1) - 1)`


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


Evaluate the following limits:

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


Choose the correct alternative:

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


Choose the correct alternative:

`lim_(x -> 3) [x]` =


Choose the correct alternative:

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


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


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