English

If ,,f(x)={x+2, x≤-1cx2,x>-1, find 'c' if limx→-1f(x) exists

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

Given, `f(x) = {{:(x + 2",",  x ≤ - 1),(cx^2",", x > -1):}`

L.H.L = `lim_(x -> -1^-) f(x)`

= `lim_(x -> -1^-) (x + 2)`

= `lim(h -> 0) (-1 - h + 2)`

= `lim_(h -> 0) (1 - h)` = 1

R.H.L = `lim_(x -> 1^+) cx^2`

= `lim_(h -> 0) c(-1 + h)^2` = c

Since the limits exist.

∴ L.H.L = R.H.L

∴ c = 1

shaalaa.com
  Is there an error in this question or solution?
Chapter 13: Limits and Derivatives - Exercise [Page 242]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 13 Limits and Derivatives
Exercise | Q 53 | Page 242

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find `lim_(x -> 5) f(x)`, where f(x)  = |x| - 5


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


\[\lim_{x \to 3} \frac{\sqrt{2x + 3}}{x + 3}\] 


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


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


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


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


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


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


If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = 9,\] find all possible values of a. 


\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\] 


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


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


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


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


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


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


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


Evaluate the following limits: 

\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\] 

 


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


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


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


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


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


\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\] 


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


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


\[\lim_{x \to 0} \frac{\sin 2x}{x}\] 


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


\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to


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


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 


Evaluate the following limit:

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


Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`


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


Let `f(x) = {{:((k cos x)/(pi - 2x)",", "when"  x ≠ pi/2),(3",", x = pi/2  "and if"  f(x) = f(pi/2)):}` find the value of k.


Evaluate the following limit:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×