Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
Advertisements
उत्तर
\[\lim_{x \to 1} \left[ \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the numerator:
\[\lim_{x \to 1} \left[ \frac{\left( \sqrt{3 + x} - \sqrt{5 - x} \right)\left( \sqrt{3 + x} + \sqrt{5 - x} \right)}{\left( x - 1 \right)\left( x + 1 \right)\left( \sqrt{3 + x} + \sqrt{5 - x} \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 3 + x \right) - \left( 5 - x \right)}{\left( x - 1 \right)\left( x + 1 \right)\left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\]
= \[\lim_{x \to 1} \left[ \frac{2x - 2}{\left( x - 1 \right)\left( x + 1 \right) \left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\]
= \[\lim_{x \to 1} \left[ \frac{2\left( x - 1 \right)}{\left( x - 1 \right)\left( x + 1 \right)\left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\]
= \[\frac{2}{2 \left( \sqrt{4} + \sqrt{4} \right)}\]
= \[\frac{1}{4}\]
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):}`
Find `lim_(x -> 1)` f(x), where `f(x) = {(x^2 -1, x <= 1), (-x^2 -1, x > 1):}`
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)`
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
\[\lim_{x \to 2} \frac{\sqrt{1 + 4x} - \sqrt{5 + 2x}}{x - 2}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - \sqrt{x + 1}}{2 x^2}\]
\[\lim_{x \to 4} \frac{2 - \sqrt{x}}{4 - x}\]
\[\lim_{x \to a} \frac{x - a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + 3x} - \sqrt{1 - 3x}}{x}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to 1} \frac{ x^2 - \sqrt{x}}{\sqrt{x} - 1}\]
\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\]
\[\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - 10}\]
\[\lim_{x \to \sqrt{2}} \frac{\sqrt{3 + 2x} - \left( \sqrt{2} + 1 \right)}{x^2 - 2}\]
\[\lim_{x \to 0} \frac{5^x - 1}{\sqrt{4 + x} - 2}\]
\[\lim_{x \to 0} \frac{a^{mx} - 1}{b^{nx} - 1}, n \neq 0\]
\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\]
\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]
\[\lim_{x \to 0} \frac{e^{2x} - e^x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log \left( a - x \right)}{x}\]
\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\]
\[\lim_{x \to 5} \frac{e^x - e^5}{x - 5}\]
\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
\[\lim_{x \to 0} \frac{\sin x}{\sqrt{1 + x} - 1} .\]
Write the value of \[\lim_{n \to \infty} \frac{n! + \left( n + 1 \right)!}{\left( n + 1 \right)! + \left( n + 2 \right)!} .\]
Write the value of \[\lim_{n \to \infty} \frac{1 + 2 + 3 + . . . + n}{n^2} .\]
Evaluate: `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 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 ______.
