English

Lim X → 3 X 2 − 4 X + 3 X 2 − 2 X − 3

Advertisements
Advertisements

Question

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

Advertisements

Solution

\[\lim_{x \to 3} \left[ \frac{x^2 - 4x + 3}{x^2 - 2x - 3} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 3} \left[ \frac{x^2 - x - 3x + 3}{x^2 - 3x + x - 3} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{\left( x - 3 \right)\left( x - 1 \right)}{\left( x + 1 \right)\left( x - 3 \right)} \right]\]
\[ = \lim_{x \to 3} \left( \frac{x - 1}{x + 1} \right)\]
\[ = \frac{3 - 1}{3 + 1}\]
\[ = \frac{1}{2}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.3 [Page 23]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.3 | Q 2 | Page 23

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}\] 


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


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\] 


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


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


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


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


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


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


Evaluate the following limit: 

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


If  \[\lim_{x \to 0} kx  cosec x = \lim_{x \to 0} x  cosec kx,\] 


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


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


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


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


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


Evaluate the following limit:

\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]

 


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


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


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


Write the value of \[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2} .\] 


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


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


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


\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] is equal to 


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


Let f(x) = `{{:(3^(1/x);   x < 0","                "then at"  x = 0),(lambda[x];   x ≥ 0","   lambda ∈ "R"):}`

Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`


`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)]`


Evaluate the following limit:

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


Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×