मराठी

Lim X → 1 X 2 + 1 X + 1 - Mathematics

Advertisements
Advertisements

प्रश्न

\[\lim_{x \to 1} \frac{x^2 + 1}{x + 1}\] 

उत्तर

\[\lim_{x \to 1} \left( \frac{x^2 + 1}{x + 1} \right)\]
\[ = \frac{1^2 + 1}{1 + 1}\]
\[ = 1\] 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.2 [पृष्ठ १८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.2 | Q 1 | पृष्ठ १८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\] 


\[\lim_{x \to 2} \frac{x^4 - 16}{x - 2}\] 


\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]


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


\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\] 


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


\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]


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


\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\] 


\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\] 


\[\lim_{x \to \infty} \frac{5 x^3 - 6}{\sqrt{9 + 4 x^6}}\]


\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\] 


\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\] 


Evaluate: \[\lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + . . . + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + . . . + n^3}{n^5}\] 


\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\] 


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


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


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


\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\] 


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


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


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


\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\] 


\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]


\[\lim_{n \to \infty} 2^{n - 1} \sin \left( \frac{a}{2^n} \right)\] 

 


\[\lim_{x \to \frac{\pi}{2}} \frac{\left( \frac{\pi}{2} - x \right) \sin x - 2 \cos x}{\left( \frac{\pi}{2} - x \right) + \cot x}\]


Evaluate the following limit:

\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]

 


\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]


\[\lim_{x \to 0} \frac{x}{\tan x} is\] 


\[\lim_{x \to 0} \frac{\sin x^0}{x}\] 


\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to 


If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]


Evaluate the following Limits: `lim_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`


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->7)[[(root3(x)- root3(7))(root3(x) + root3(7)))/(x-7)]`


Evaluate the Following limit: 

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×