English

Lim X → − 1 ( 4 X 2 + 2 )

Advertisements
Advertisements

Question

\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]

Advertisements

Solution

\[\lim_{x \to - 1} \left( 4 x^2 + 2 \right)\]
\[ = 4 \left( - 1 \right)^2 + 2\]
\[ = 4 + 2\]
\[ = 6\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.2 [Page 18]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.2 | Q 10 | Page 18

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.


\[\lim_{x \to 0} \frac{2 x^2 + 3x + 4}{x^2 + 3x + 2}\] 


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


\[\lim_{x \to 0} 9\] 


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


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


\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\] 


If \[\lim_{x \to 3} \frac{x^n - 3^n}{x - 3} = 108,\]  find the value of n


\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\] 


\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\] 


\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]


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


Evaluate: \[\lim_{n \to \infty} \frac{1 . 2 + 2 . 3 + 3 . 4 + . . . + n\left( n + 1 \right)}{n^3}\] 


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


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


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


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


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


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


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


\[\lim_\theta \to 0 \frac{\sin 4\theta}{\tan 3\theta}\] 


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


\[\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{\frac{\pi}{2} - x}\]


\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]


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


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


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


Write the value of \[\lim_{x \to \infty} \frac{\sin x}{x} .\] 


If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\]  then \[\lim_{x \to 0} f\left( x \right) =\] 


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


\[\lim 2_{h \to 0} \left\{ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h \left( \sqrt{3} \cos h - \sin h \right)} \right\}\] 


\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\]  is equal to


If `lim_(x -> 1) (x^4 - 1)/(x - 1) = lim_(x -> k) (x^3 - l^3)/(x^2 - k^2)`, then find the value of k.


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×