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 -> 1)` f(x), where `f(x) = {(x^2 -1, x <= 1), (-x^2 -1, x > 1):}`
Find `lim_(x -> 0)` f(x), where `f(x) = {(x/|x|, x != 0),(0, x = 0):}`
If f(x) = `{(|x| + 1,x < 0), (0, x = 0),(|x| -1, x > 0):}`
For what value (s) of a does `lim_(x -> a)` f(x) exists?
if `f(x) = { (mx^2 + n, x < 0),(nx + m, 0<= x <= 1),(nx^3 + m, x > 1):}`
For what integers m and n does `lim_(x-> 0) f(x)` and `lim_(x -> 1) f(x)` exist?
\[\lim_{x \to 2} \frac{\sqrt{3 - x} - 1}{2 - x}\]
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
\[\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x - 1}\]
\[\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 3 - 2}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - \sqrt{x + 1}}{2 x^2}\]
\[\lim_{x \to a} \frac{x - a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}\]
\[\lim_{x \to 1} \frac{ x^2 - \sqrt{x}}{\sqrt{x} - 1}\]
\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\]
\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]
\[\lim_{x \to 0} \frac{a^x + a^{- x} - 2}{x^2}\]
\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]
\[\lim_{x \to \infty} \left( a^{1/x} - 1 \right)x\]
\[\lim_{x \to 0} \frac{e^x - 1 + \sin x}{x}\]
\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\]
`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`
\[\lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x}\]
\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\]
\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]
\[\lim_{x \to 0} \frac{x\left( e^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to \infty} \left\{ \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right\}^\frac{3x - 2}{3x + 2}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
Evaluate: `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 2))`
