हिंदी

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

Advertisements
Advertisements

प्रश्न

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

Advertisements

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

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


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


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


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


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


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


If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = \lim_{x \to 5} \left( 4 + x \right),\] find all possible values of a


\[\lim_{x \to \infty} \frac{3 x^3 - 4 x^2 + 6x - 1}{2 x^3 + x^2 - 5x + 7}\] 


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


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


\[\lim_{n \to \infty} \left[ \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!} \right]\] 


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal


\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, is equal to 


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


Evaluate the following limit:

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


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 ______.


If f(x) = `{{:(1 if x  "is rational"),(-1 if x  "is rational"):}` is continuous on ______.


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


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×