English

Lim X → − 1 / 2 8 X 3 + 1 2 X + 1 - Mathematics

Advertisements
Advertisements

Question

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

Advertisements

Solution

\[\lim_{x \to - \frac{1}{2}} \left[ \frac{8 x^3 + 1}{2x + 1} \right]\]
\[ = \lim_{x \to - \frac{1}{2}} \left[ \frac{\left( 2x \right)^3 - \left( - 1 \right)}{2x - \left( - 1 \right)} \right]\]
\[\text{ When } x \to - \frac{1}{2}, \text{ then } 2x \to -1 .\] 

Let y = 2

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.5 [Page 33]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.5 | Q 6 | Page 33

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


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


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


\[\lim_{x \to 27} \frac{\left( x^{1/3} + 3 \right) \left( x^{1/3} - 3 \right)}{x - 27}\] 


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


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


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


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


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


\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\] 


\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\] 


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


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


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


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


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


\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\] 


Evaluate the following limit: 

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


\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\] 


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


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


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


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


Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]

 

\[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\] 


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


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


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


\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]


The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\]  is


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


Evaluate the following limits: `lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`


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


Evaluate the following limit :

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


Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`


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×