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Choose the correct alternative: limx→∞(x2+5x+3x2+x+3)x is - Mathematics

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प्रश्न

Choose the correct alternative:

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

पर्याय

  • e4

  • e2

  • e3

  • 1

MCQ
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उत्तर

e4 

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

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.6 | Q 5 | पृष्ठ १२९

संबंधित प्रश्‍न

Evaluate the following limit:

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


Evaluate the following limit:

`lim_(x -> 3)[sqrt(2x + 6)/x]`


Evaluate the following limit:

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

Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(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 -> 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) sin x/x`

x – 0.1  – 0.01 – 0.001 0.001 0.01 0.1
f(x) 0.99833 0.99998 0.99999 0.99999 0.99998 0.99833

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 -> 5) |x - 5|/(x - 5)`


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) sin pi x`


Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.

f(x) = `{{:(x^2",", x ≤ 2),(8 - 2x",", 2 < x < 4),(4",", x ≥ 4):}`


Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.

f(x) = `{{:(sin x",", x < 0),(1 - cos x",", 0 ≤ x ≤ pi),(cos x",", x > pi):}`


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 -> 2) (1/x - 1/2)/(x - 2)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Choose the correct alternative:

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


Choose the correct alternative:

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


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


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