Advertisements
Advertisements
Question
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2x}\]
Advertisements
Solution
\[\lim_{x \to 0} \left[ \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2x} \right]\] It is of the form\[\frac{0}{0} .\]
Rationalising the numerator:
\[\lim_{x \to 0} \left[ \frac{\left( \sqrt{1 + x} - \sqrt{1 - x} \right)\left( \sqrt{1 + x} + \sqrt{1 - x} \right)}{2x\left( \sqrt{1 + x} + \sqrt{1 - x} \right)} \right]\]
\[\lim_{x \to 0} \left[ \frac{\left( 1 + x \right) - \left( 1 - x \right)}{2x\left( \sqrt{1 + x} + \sqrt{1 - x} \right)} \right]\]
\[\lim_{x \to 0} \left[ \frac{2x}{2x\left( \sqrt{1 + x} + \sqrt{1 - x} \right)} \right]\]
\[\frac{1}{\sqrt{1 + 0} + \sqrt{1 - 0}}\]
\[\frac{1}{2}\]
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):}`
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 the function f(x) satisfies `lim_(x -> 1) (f(x) - 2)/(x^2 - 1) = pi`, evaluate `lim_(x -> 1) f(x)`.
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\]
\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\]
\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]
\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\]
\[\lim_{x \to a} \frac{x - a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}\]
\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\]
\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\]
\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]
\[\lim_{x \to 0} \frac{a^{mx} - 1}{b^{nx} - 1}, n \neq 0\]
\[\lim_{x \to 0} \frac{9^x - 2 . 6^x + 4^x}{x^2}\]
\[\lim_{x \to \infty} \left( a^{1/x} - 1 \right)x\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{\sin kx}\]
\[\lim_{x \to 0} \frac{a^x + b^ x - c^x - d^x}{x}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\]
\[\lim_{x \to 0} \frac{x\left( 2^x - 1 \right)}{1 - \cos x}\]
`\lim_{x \to \pi/2} \frac{a^\cot x - a^\cos x}{\cot x - \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}`
\[\lim_{x \to 0} \frac{x\left( e^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
Write the value of \[\lim_{n \to \infty} \frac{n! + \left( n + 1 \right)!}{\left( n + 1 \right)! + \left( n + 2 \right)!} .\]
Evaluate: `lim_(h -> 0) (sqrt(x + h) - sqrt(x))/h`
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 ______.
