Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 2} \frac{\sqrt{3 - x} - 1}{2 - x}\]
Advertisements
उत्तर
\[\lim_{x \to 2} \left[ \frac{\sqrt{3 - x} - 1}{2 - x} \right]\] It is of the form \[\frac{0}{0} .\]Rationalising the numerator:
\[\lim_{x \to 2} \left[ \frac{\left( \sqrt{3 - x} - 1 \right)\left( \sqrt{3 - x} + 1 \right)}{\left( 2 - x \right)\left( \sqrt{3 - x} + 1 \right)} \right]\]
\[= \lim_{x \to 2} \left[ \frac{3 - x - 1}{\left( 2 - x \right)\left( \sqrt{3 - x} + 1 \right)} \right]\]
\[= \lim_{x \to 2} \left[ \frac{\left( 2 - x \right)}{\left( 2 - x \right)\left( \sqrt{3 - x} + 1 \right)} \right]\]
\[= \frac{1}{\sqrt{3 - 2} + 1}\]
\[ = \frac{1}{1 + 1}\]
\[ = \frac{1}{2}\]
APPEARS IN
संबंधित 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):}`
Evaluate `lim_(x -> 0) f(x)` where `f(x) = { (|x|/x, x != 0),(0, x = 0):}`
If the function f(x) satisfies `lim_(x -> 1) (f(x) - 2)/(x^2 - 1) = pi`, evaluate `lim_(x -> 1) f(x)`.
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - 1}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\]
\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\]
\[\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}\]
\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\]
\[\lim_{x \to a} \frac{x - a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}\]
\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]
\[\lim_{x \to 0} \frac{a^{mx} - 1}{b^{nx} - 1}, n \neq 0\]
\[\lim_{x \to 0} \frac{a^x + b^x - 2}{x}\]
\[\lim_{x \to 0} \frac{a^x + b^x + c^x - 3}{x}\]
\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]
\[\lim_{x \to \infty} \left( a^{1/x} - 1 \right)x\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{\sin kx}\]
\[\lim_{x \to 0} \frac{\log \left( 2 + x \right) + \log 0 . 5}{x}\]
\[\lim_{x \to 5} \frac{e^x - e^5}{x - 5}\]
\[\lim_{x \to 0} \frac{e^{3 + x} - \sin x - e^3}{x}\]
\[\lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x}\]
\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
\[\lim_{x \to 0} \frac{\sin x}{\sqrt{1 + x} - 1} .\]
Write the value of \[\lim_{x \to - \infty} \left( 3x + \sqrt{9 x^2 - x} \right) .\]
Write the value of \[\lim_{n \to \infty} \frac{n! + \left( n + 1 \right)!}{\left( n + 1 \right)! + \left( n + 2 \right)!} .\]
Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If `lim_(x rightarrow 0) ((f(x))/x^2 + 1)` = 3 then f(–1) is equal to ______.
