हिंदी

Evaluate the following Limits: limx→a(x+2)53-(a+2)53x-a

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

`lim_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`

Put x + 2 = y and a + 2 = b
As x → a, x + 2 → a + 2
i.e. y → b.

∴ `lim_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`

= `lim_(x -> "a") (y^(5/3) - "b"^(5/3))/((y - 2) - ("b" - 2)`

= `lim_(y -> "b") (y^(5/3) - "b"^(5/3))/(y - "b")`

= `5/3"b"^(2/3)   ...[lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`

= `5/3("a" + 2)^(2/3)`    ...[∵ b = a + 2]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - MISCELLANEOUS EXERCISE - 7 [पृष्ठ १०५]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
MISCELLANEOUS EXERCISE - 7 | Q II. 1) | पृष्ठ १०५

संबंधित प्रश्न

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


\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\] 


\[\lim_{x \to 5} \frac{x^2 - 9x + 20}{x^2 - 6x + 5}\] 


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


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


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


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


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


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


\[\lim_{x \to \frac{\pi}{4}} \frac{2 - {cosec}^2 x}{1 - \cot x}\] 


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


\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\] 


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


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


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


\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to 


If \[f\left( x \right) = \left\{ \begin{array}{l}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0\end{array}, \right.\] then \[\lim_{x \to 0} f\left( x \right)\]  equals 


\[\lim_\theta \to \pi/2 \frac{1 - \sin \theta}{\left( \pi/2 - \theta \right) \cos \theta}\] is equal to 


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


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×