हिंदी

Find limx→0 f(x) and limx→1 f(x) where f(x) = {2x+3x≤03(x+1)x>0

Advertisements
Advertisements

प्रश्न

Find `lim_(x -> 0)` f(x) and `lim_(x -> 1)` f(x) where f(x) = `{(2x + 3, x <= 0),(3(x+1), x > 0):}`

योग
Advertisements

उत्तर

`f(x) = {(2x + 3, x ≤ 0),(3(x+1), x > 0):}`

`lim_(x → 0^-) f(x) = lim_(x → 0)[2x + 3] = 2(0) + 3 = 3`

`lim_(x → 0^+) f(x) = lim_(x → 0) 3(x + 1) = 3(0 + 1) = 3`

∴ `lim_(x → 0^-) f(x) = lim_(x → 0^+) f(x) = lim_(x → 0) f(x) = 3`

`lim_(x → 1^-) f(x) = lim_(x → 1) 3(x + 1) = 3(1 + 1) = 6`

`lim_(x → 1^+) f(x) = lim_(x → 1) 3(x + 1) = 3(1 + 1) = 6`

∴ `lim_(x → 1) f(x) = lim_(x → 1^-) f(x) = lim_(x → 1^+) f(x) = 6`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Limits and Derivatives - EXERCISE 12.1 [पृष्ठ २३८]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 11
अध्याय 12 Limits and Derivatives
EXERCISE 12.1 | Q 23. | पृष्ठ २३८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

If f(x) = `{(|x| +  1,x < 0), (0, x = 0),(|x| -1, x > 0):}`

For what value (s) of a does `lim_(x -> a)`  f(x) exists?


If the function f(x) satisfies `lim_(x -> 1) (f(x) - 2)/(x^2 - 1) = pi`, evaluate `lim_(x -> 1) f(x)`.


if `f(x) = { (mx^2 + n, x < 0),(nx + m, 0<= x <= 1),(nx^3 + m, x > 1):}`

For what integers m and n does `lim_(x-> 0) f(x)` and `lim_(x -> 1) f(x)` exist?


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


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


\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x - 1}\] 


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


\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\] 


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


\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]


\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\] 


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


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


\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}\] 


\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\] 

 


\[\lim_{x \to \sqrt{2}} \frac{\sqrt{3 + 2x} - \left( \sqrt{2} + 1 \right)}{x^2 - 2}\] 


\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\] 


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


\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]


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


\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\] 


`\lim_{x \to \pi/2} \frac{a^\cot x - a^\cos x}{\cot x - \cos x}`


\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]


\[\lim_{x \to 5} \frac{e^x - e^5}{x - 5}\]


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


`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`


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


`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`


`\lim_{x \to 0} \frac{e^x - e^\sin x}{x - \sin x}`


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


\[\lim_{x \to \infty} \left\{ \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right\}^\frac{3x - 2}{3x + 2}\]


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


\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]


Evaluate: `lim_(h -> 0) (sqrt(x + h) - sqrt(x))/h`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×