हिंदी

Lim X → ∞ | X | X is Equal to - Mathematics

Advertisements
Advertisements

प्रश्न

\[\lim_{x \to \infty} \frac{\left| x \right|}{x}\]  is equal to 

विकल्प

  •  1     

  • −1         

  •  0     

  •  does not exist 

MCQ
Advertisements

उत्तर

We know that, 

\[\left| x \right| = \begin{cases}x, & if x \geq 0 \\ - x, & if x < 0\end{cases}\]
\[ \therefore \frac{\left| x \right|}{x} = \begin{cases}\frac{x}{x}, & if x \geq 0 \\ \frac{- x}{x}, & if x < 0\end{cases} = \begin{cases}1, & if x \geq 0 \\ - 1, & if x < 0\end{cases}\]
Now, for all  0 (however, x may large be),  

\[\frac{\left| x \right|}{x} = 1\] 

\[\therefore \lim_{x \to \infty} \frac{\left| x \right|}{x} = 1\] 

Hence, the correct answer is option (a).

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.13 [पृष्ठ ८१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.13 | Q 40 | पृष्ठ ८१

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\] 


\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\] 


\[\lim_{x \to 2} \left( \frac{x}{x - 2} - \frac{4}{x^2 - 2x} \right)\] 


\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]


\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\] 


Evaluate the following limit:

\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\] 


\[\lim_{x \to a} \frac{x^{2/7} - a^{2/7}}{x - a}\] 


If \[\lim_{x \to a} \frac{x^3 - a^3}{x - a} = \lim_{x \to 1} \frac{x^4 - 1}{x - 1},\] find all possible values of a


`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`


\[\lim_{n \to \infty} \left[ \frac{1 + 2 + 3 . . . . . . n - 1}{n^2} \right]\] 


\[\lim_{x \to 0} \frac{\sin 3x}{5x}\] 


\[\lim_{x \to 0} \frac{\sin x \cos x}{3x}\] 


\[\lim_{x \to 0} \frac{\tan^2 3x}{x^2}\] 


\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\] 


\[\lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x}\]


\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\] 


\[\lim_{x \to 0} \frac{2 \sin x^\circ - \sin 2 x^\circ}{x^3}\] 


Evaluate the following limits: 

\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}\] 


Evaluate the following limit: 

\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\] 


Evaluate the following limit:

\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]


\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\] 


\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\] 


\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\] 


\[\lim_{x \to 1} \left( 1 - x \right) \tan \left( \frac{\pi x}{2} \right)\]


\[\lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{cosec x - 2}\]


\[\lim_{x \to \frac{\pi}{4}} \frac{2 - {cosec}^2 x}{1 - \cot x}\] 


\[\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + . . . + n^2}{n^3}\] 


\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to


\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to


The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\] 


Evaluate the following limit:

`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`


Evaluate the following limits: if `lim_(x -> 1)[(x^4 - 1)/(x - 1)] = lim_(x -> "a") [(x^3 - "a"^3)/(x - "a")]`, find all the value of a.


Evaluate the following limit:

`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`


Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`


if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.


Let f(x) = `{{:(3^(1/x);   x < 0","                "then at"  x = 0),(lambda[x];   x ≥ 0","   lambda ∈ "R"):}`

If `f(x) = {{:(x + 2",",  x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists


Number of values of x where the function

f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3","  π):}`

is discontinuous, is ______.


Evaluate the following limit:

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


Evaluate the following limit:

`lim _ (x -> 5) [(x^3 - 125) / (x^5 - 3125)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×