मराठी

If Lim X → a X 3 − a 3 X − a = Lim X → 1 X 4 − 1 X − 1 , Find All Possible Values of A. - Mathematics

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to a} \left[ \frac{x^3 - a^3}{x - a} \right] = \lim_{x \to 1} \left[ \frac{x^4 - 1^4}{x - 1} \right]\]
\[ \Rightarrow 3 a^{3 - 1} = 4 \left( 1 \right)^{4 - 1} \]
\[ \Rightarrow 3 a^2 = 4\]
\[ \Rightarrow a^2 = \frac{4}{3}\]
\[ \Rightarrow a = \pm \frac{2}{\sqrt{3}}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.5 [पृष्ठ ३३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.5 | Q 16 | पृष्ठ ३३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\] 


\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\] 


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


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


\[\lim_{x \to - 2} \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2}\]


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


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


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->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]


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


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


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


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


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


Evaluate the following limit: 

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


\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\] 


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


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


\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]


\[\lim_{x \to 0}  \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]


Evaluate the following Limit:

`lim_(x -> 0) ((1 + x)^"n" - 1)/x`


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.


Evaluate the Following limit:

`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×