Advertisements
Advertisements
Question
Evaluate the following one sided limit:
\[\lim_{x \to 2^-} \frac{x - 3}{x^2 - 4}\]
Solution
\[\lim_{x \to 2^-} \left( \frac{x - 3}{x^2 - 4} \right)\]
\[\text{ Let } x = 2 - h, \text{ where } h \to 0 . \]
\[ \Rightarrow \lim_{h \to 0} \left[ \frac{2 - h - 3}{\left( 2 - h \right)^2 - 2^2} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{- h - 1}{\left( 2 - h - 2 \right) \left( 2 - h + 2 \right)} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{- h - 1}{\left( - h \right) \left( 4 - h \right)} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{1 + h}{h\left( 4 - h \right)} \right]\]
\[ = \infty\]
APPEARS IN
RELATED QUESTIONS
Show that \[\lim_{x \to 0} \frac{1}{x}\] does not exist.
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 \[f\left( x \right) = \begin{cases}x + 5, & if x > 0 \\ x - 4, & if x < 0\end{cases}\] \[\lim_{x \to 0} f\left( x \right)\] 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 \[\lim_{x \to 1} f\left( x \right)\]
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 \[f\left( x \right) = \begin{cases}\frac{\left| x \right|}{x}, & x \neq 0 \\ 0, & x = 0\end{cases}\]
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:
\[\lim_{x \to 0^+} \frac{2}{x^{1/5}}\]
Evaluate the following one sided limit:
\[\lim_{x \to \frac{\pi}{2}} \tan x\]
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:
\[\lim_{x \to 0^-} 2 - \cot x\]
Evaluate the following one sided limit:
\[\lim_{x \to 0^-} 1 + cosec x\]
Show that \[\lim_{x \to 0} e^{- 1/x}\] does not exist.
Find: \[\ \lim_{x \to 2} \left[ x \right]\]
Find: \[ \lim_{x \to \frac{5}{2}} \left[ x \right]\]
Find: \[ \lim_{x \to 1} \left[ x \right]\]
Prove that \[\lim_{x \to a^+} \left[ x \right] = \left[ a \right]\] R. Also, prove that \[\lim_{x \to 1^-} \left[ x \right] = 0 .\]
Show that \[\lim_{x \to 2^-} \frac{x}{\left[ x \right]} \neq \lim_{x \to 2^+} \frac{x}{\left[ x \right]} .\]
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 \[\lim_{x \to \frac{\pi}{2}} f\left( x \right) = f\left( \frac{\pi}{2} \right)\]
