Advertisements
Advertisements
Question
Write the value of \[\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x} .\]
Advertisements
Solution
\[\lim_{x \to \frac{\pi}{2}} \left[ \frac{2x - \pi}{\cos x} \right]\]
\[LHL: \]
\[ \lim_{x \to \frac{\pi}{2}^-} \left[ \frac{2x - \pi}{\cos x} \right]\]
\[Let x = \frac{\pi}{2} - h\]
\[\text{ If } x \to \frac{\pi}{2}, \text{ then we have }: \]
\[ h \to 0\]
\[ = \lim_{h \to 0} \left[ \frac{2\left( \frac{\pi}{2} - h \right) - \pi}{\cos \left( \frac{\pi}{2} - h \right)} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{\pi - 2h - \pi}{\sin h} \right]\]
\[ = - 2\]
\[RHL: \]
\[ \lim_{x \to \frac{\pi}{2}^+} \left[ \frac{2x - \pi}{\cos x} \right]\]
\[\text{ Let } x = \frac{\pi}{2} + h\]
\[\text{ If } x \to \frac{\pi}{2}, \text{ then we have }: \]
\[h \to 0\]
\[ = \lim_{h \to 0} \left[ \frac{2\left( \frac{\pi}{2} + h \right) - \pi}{\cos \left( \frac{\pi}{2} + h \right)} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{2h}{- \sin h} \right]\]
\[ = - 2\]
\[ \Rightarrow \lim_{x \to \frac{\pi}{2}} \left( \frac{2x - \pi}{\cos x} \right) = - 2\]
APPEARS IN
RELATED QUESTIONS
Find `lim_(x -> 0)` f(x) and `lim_(x -> 1)` f(x) where f(x) = `{(2x + 3, x <= 0),(3(x+1), x > 0):}`
Find `lim_(x -> 0)` f(x), where `f(x) = {(x/|x|, x != 0),(0, x = 0):}`
Let a1, a2,..., an be fixed real numbers and define a function f ( x) = ( x − a1 ) ( x − a2 )...( x − an ).
What is `lim_(x -> a_1) f(x)` ? For some a ≠ a1, a2, ..., an, compute `lim_(x -> a) f(x)`
If the function f(x) satisfies `lim_(x -> 1) (f(x) - 2)/(x^2 - 1) = pi`, evaluate `lim_(x -> 1) f(x)`.
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2x}\]
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
\[\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\]
\[\lim_{x \to 2} \frac{\sqrt{1 + 4x} - \sqrt{5 + 2x}}{x - 2}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\]
\[\lim_{x \to 4} \frac{2 - \sqrt{x}}{4 - x}\]
\[\lim_{x \to a} \frac{x - a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to 0} \frac{\sqrt{2 - x} - \sqrt{2 + x}}{x}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]
\[\lim_{x \to 0} \frac{a^x + b^x - 2}{x}\]
\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]
\[\lim_{x \to 0} \frac{a^x + b^ x - c^x - d^x}{x}\]
\[\lim_{x \to 0} \frac{e\sin x - 1}{x}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log \left( a - x \right)}{x}\]
\[\lim_{x \to 0} \frac{\log \left( 2 + x \right) + \log 0 . 5}{x}\]
\[\lim_{x \to 0} \frac{x\left( 2^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\]
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
\[\lim_{x \to 0} \frac{e^{x + 2} - e^2}{x}\]
\[\lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x}\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{\tan x}`
`\lim_{x \to 0} \frac{e^x - e^\sin x}{x - \sin x}`
\[\lim_{x \to \infty} \left\{ \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right\}^\frac{3x - 2}{3x + 2}\]
\[\lim_{x \to 1} \left\{ \frac{x^3 + 2 x^2 + x + 1}{x^2 + 2x + 3} \right\}^\frac{1 - \cos \left( x - 1 \right)}{\left( x - 1 \right)^2}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
Write the value of \[\lim_{n \to \infty} \frac{1 + 2 + 3 + . . . + n}{n^2} .\]
