Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}\]
Advertisements
उत्तर
\[\lim_{x \to 0} \left[ \frac{\sqrt{1 - x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the numerator and the denominator:
\[\lim_{x \to 0} \left[ \frac{\left( \sqrt{1 + x^2} - \sqrt{1 + x} \right)}{1} \times \frac{\left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)}{\left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)} \times \frac{1}{\left( \sqrt{1 + x^3} - \sqrt{1 + x} \right)} \times \frac{\left( \sqrt{1 + x^3} - \sqrt{1 + x} \right)}{\left( \sqrt{1 + x^3} + \sqrt{1 + x} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left[ \left( 1 + x^2 \right) - \left( 1 + x \right) \right]}{\left[ \left( 1 + x^3 \right) - \left( 1 + x \right) \right]} \times \frac{\left( \sqrt{1 + x^3} + \sqrt{1 + x} \right)}{\left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \left( \frac{x^2 - x}{x^3 - x} \right) \times \frac{\left( \sqrt{1 + x^3} + \sqrt{1 + x} \right)}{\left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{x\left( x - 1 \right)}{x\left( x^2 - 1 \right)} \frac{\left( \sqrt{1 + x^3} + \sqrt{1 + x} \right)}{\left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( x - 1 \right) \left( \sqrt{1 + x^3} + \sqrt{1 + x} \right)}{\left( x - 1 \right) \left( x + 1 \right) \left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)} \right]\]
\[ = \frac{\left( \sqrt{1 + 0} + \sqrt{1 + 0} \right)}{\left( 0 + 1 \right) \left( \sqrt{1 + 0} + \sqrt{1 + 0} \right)}\]
\[ = \frac{2}{2}\]
\[ = 1\]
APPEARS IN
संबंधित प्रश्न
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)`
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - 1}{x}\]
\[\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\]
\[\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 3 - 2}}\]
\[\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1}\]
\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\]
\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\]
\[\lim_{x \to a} \frac{x - a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - 10}\]
\[\lim_{x \to 0} \frac{a^x + b^x - 2}{x}\]
\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]
\[\lim_{x \to \infty} \left( a^{1/x} - 1 \right)x\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{\sin kx}\]
\[\lim_{x \to 0} \frac{a^x + b^ x - c^x - d^x}{x}\]
\[\lim_{x \to 0} \frac{e^{2x} - e^x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{\log \left( 2 + x \right) + \log 0 . 5}{x}\]
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`
`\lim_{x \to 0} \frac{e^\tan x - 1}{\tan x}`
\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`
`\lim_{x \to 0} \frac{e^x - e^\sin x}{x - \sin x}`
\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]
\[\lim_{x \to \infty} \left\{ \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right\}^\frac{3x - 2}{3x + 2}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
Write the value of \[\lim_{n \to \infty} \frac{n! + \left( n + 1 \right)!}{\left( n + 1 \right)! + \left( n + 2 \right)!} .\]
Write the value of \[\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x} .\]
Write the value of \[\lim_{n \to \infty} \frac{1 + 2 + 3 + . . . + n}{n^2} .\]
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 ______.
