हिंदी

If α is a Repeated Root of Ax2 + Bx + C = 0, Then Lim X → α Tan ( a X 2 + B X + C ) ( X − α ) 2 - Mathematics

Advertisements
Advertisements

प्रश्न

If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]

विकल्प

  •  

  •  

  •  0

MCQ
Advertisements

उत्तर

\[\lim_{x \to \alpha} \left[ \frac{\tan\left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2} \right]\]
\[ = \lim_{x \to \alpha} \left[ \frac{\tan\left\{ a\left( x - \alpha \right)\left( x - \alpha \right) \right\}}{a \left( x - \alpha \right)^2} \right] \times a\]
\[ = \lim_{x \to \alpha} \left[ \frac{\tan\left\{ a \left( x - \alpha \right)^2 \right\}}{a \left( x - \alpha \right)^2} \right] \times a\]
\[ = 1 \times a \left[ \lim_\theta \to 0 \left( \frac{\tan\theta}{\theta} \right) = 1 \right]\]
\[ = a\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.13 [पृष्ठ ८०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.13 | Q 30 | पृष्ठ ८०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Suppose f(x)  = `{(a+bx, x < 1),(4, x = 1),(b-ax, x > 1):}`  and if `lim_(x -> 1) f(x) = f(1)` what are possible values of a and b?


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


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


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


\[\lim_{x \to 5} \frac{x^3 - 125}{x^2 - 7x + 10}\] 


\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\] 


\[\lim_{x \to \infty} \frac{5 x^3 - 6}{\sqrt{9 + 4 x^6}}\]


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x}\] 


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


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


\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]


\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\] 


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


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


\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}\] 


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


\[\lim_{x \to 0} \frac{\log \left( 3 + x \right) - \log \left( 3 - x \right)}{x}\] 


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


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


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


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


\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to 


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


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


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


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


Evaluate the following limit:

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


Evaluate the following Limit:

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


If `f(x) = {{:(x + 2",",  x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists


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 ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×