मराठी

Find: Lim X → 5 2 [ X ] - Mathematics

Advertisements
Advertisements

प्रश्न

Find: \[ \lim_{x \to \frac{5}{2}} \left[ x \right]\]

 

उत्तर

\[\lim_{x \to \frac{5}{2}} \left[ x \right]\]
\[\text{ LHL }\]
\[ \lim_{x \to \frac{5}{2}^-} \left[ x \right]\]
\[\text{ Let } x = \frac{5}{2} - \text{ h, where h } \to 0 . \]
\[ \lim_{h \to 0} \left[ \frac{5}{2} - h \right]\]
\[ = 2\]
\[\text{ RHL }: \]
\[ \lim_{x \to \frac{5}{2}^+} \left[ x \right]\]
\[\text{ Let } x = \frac{5}{2} + \text{ h, where h } \to 0 . \]
\[ \Rightarrow \lim_{h \to 0} \left[ \frac{5}{2} + h \right] \]
\[ \therefore \lim_{x \to \frac{5}{2}} \left[ x \right] = 2\]
\[ = 2\]
\[ \therefore \lim_{x \to \frac{5}{2}} \left[ x \right] = 2\]

 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.1 [पृष्ठ १२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.1 | Q 15.2 | पृष्ठ १२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Let f(x) be a function defined by \[f\left( x \right) = \begin{cases}\frac{3x}{\left| x \right| + 2x}, & x \neq 0 \\ 0, & x = 0\end{cases} .\] Show that \[\lim_{x \to 0} f\left( x \right)\] does not exist.

 

Let \[f\left( x \right) = \left\{ \begin{array}{l}x + 1, & if x \geq 0 \\ x - 1, & if x < 0\end{array} . \right.\]Prove that \[\lim_{x \to 0} f\left( x \right)\] does not exist.


Let \[f\left( x \right) = \begin{cases}x + 5, & if x > 0 \\ x - 4, & if x < 0\end{cases}\] \[\lim_{x \to 0} f\left( x \right)\]  does not exist. 


Find \[\lim_{x \to 3} f\left( x \right)\] where \[f\left( x \right) = \begin{cases}4, & if x > 3 \\ x + 1, & if x < 3\end{cases}\] 


If \[f\left( x \right) = \left\{ \begin{array}{l}2x + 3, & x \leq 0 \\ 3 \left( x + 1 \right), & x > 0\end{array} . \right.\] find \[\lim_{x \to 0} f\left( x \right)\] 


If \[f\left(  x \right) = \left\{ \begin{array}{l}2x + 3, & x \leq 0 \\ 3 \left( x + 1 \right), & x > 0\end{array} . \right.\] find \[\lim_{x \to 1} f\left( x \right)\]


Find \[\lim_{x \to 1} f\left( x \right)\] if \[f\left( x \right) = \begin{cases}x^2 - 1, & x \leq 1 \\ - x^2 - 1, & x > 1\end{cases}\] 


Evaluate \[\lim_{x \to 0} f\left( x \right)\]  where \[f\left( x \right) = \begin{cases}\frac{\left| x \right|}{x}, & x \neq 0 \\ 0, & x = 0\end{cases}\] 


Let a1a2, ..., an be fixed real numbers such that
f(x) = (x − a1) (x − a2) ... (x − an)
What is \[\lim_{x \to a_1} f\left( x \right)?\] Compute \[\lim_{x \to a} f\left( x \right) .\] 


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


Evaluate the following one sided limit: 

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


Evaluate the following one sided limit: 

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


Evaluate the following one sided limit:

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


Evaluate the following one sided limit: 

\[\lim_{x \to \frac{\pi}{2}} \tan x\]


Evaluate the following one sided limit:

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


Evaluate the following one sided limit:

\[\lim_{x \to 0^-} 2 - \cot x\] 


Show that \[\lim_{x \to 0} e^{- 1/x}\] does not exist. 


Prove that \[\lim_{x \to a^+} \left[ x \right] = \left[ a \right]\] R. Also, prove that \[\lim_{x \to 1^-} \left[ x \right] = 0 .\]


Find \[\lim_{x \to 5/2} \left[ x \right] .\] 


Evaluate \[\lim_{x \to 2} f\left( x \right)\] (if it exists), where \[f\left( x \right) = \left\{ \begin{array}{l}x - \left[ x \right], & x < 2 \\ 4, & x = 2 \\ 3x - 5, & x > 2\end{array} . \right.\]


Let \[f\left( x \right) = \begin{cases}\frac{k\cos x}{\pi - 2x}, & where x \neq \frac{\pi}{2} \\ 3, & where x = \frac{\pi}{2}\end{cases}\]   and if \[\lim_{x \to \frac{\pi}{2}} f\left( x \right) = f\left( \frac{\pi}{2} \right)\] 


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

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

\[\lim_{n \to \infty} n \sin \left( \frac{\pi}{4 n} \right) \cos \left( \frac{\pi}{4 n} \right)\]

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×