हिंदी

lim x → 2 x 4 − 16 x − 2 - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\lim_{x \to 2} \left[ \frac{x^4 - 16}{x - 2} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 2} \left[ \frac{\left( x^2 \right)^2 - \left( 4 \right)^2}{x - 2} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{\left( x^2 - 4 \right)\left( x^2 + 4 \right)}{x - 2} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{\left( x - 2 \right)\left( x + 2 \right)\left( x^2 + 4 \right)}{x - 2} \right]\]
\[ = \left( 2 + 2 \right)\left( 2^2 + 4 \right)\]
\[ = 32\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.3 [पृष्ठ २३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.3 | Q 7 | पृष्ठ २३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\lim_{x \to 0} 9\] 


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


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


\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\] 


\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]


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


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


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


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


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


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


\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number. 


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


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


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


\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]


\[\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}\]


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


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


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


\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]


Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]


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


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


\[\lim_{x \to \infty} \frac{\sin x}{x} .\] 


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


\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\]  is equal at 


The value of \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is 


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


\[\lim_{x \to 0} \frac{\left| \sin x \right|}{x}\]


Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`


Evaluate the following limits: `lim_(y -> 1) [(2y - 2)/(root(3)(7 + y) - 2)]`


Which of the following function is not continuous at x = 0?


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.


`1/(ax^2 + bx + c)`


Evaluate the following limit:

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


Evaluate the following limit:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×