Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 3 - 2}}\]
Advertisements
उत्तर
\[\lim_{x \to 1} \left[ \frac{x - 1}{\sqrt{x^2 + 3} - 2} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the denominator:
\[\lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( \sqrt{x^2 + 3} + 2 \right)}{\left( \sqrt{x^2 + 3} - 2 \right)\left( \sqrt{x^2 + 3} + 2 \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( \sqrt{x^2 + 3} + 2 \right)}{x^2 + 3 - 4} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( \sqrt{x^2 + 3} + 2 \right)}{\left( x^2 - 1 \right)} \right]\] = \[\lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( \sqrt{x^2 + 3} + 2 \right)}{\left( x - 1 \right)\left( x + 1 \right)} \right]\]
= \[\frac{\sqrt{1 + 3} + 2}{1 + 1}\]
= \[\frac{4}{2}\]
= 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):}`
If the function f(x) satisfies `lim_(x -> 1) (f(x) - 2)/(x^2 - 1) = pi`, evaluate `lim_(x -> 1) f(x)`.
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?
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - 1}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{a^2 + x^2} - a}{x^2}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\]
\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\]
\[\lim_{x \to 2} \frac{\sqrt{1 + 4x} - \sqrt{5 + 2x}}{x - 2}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - \sqrt{x + 1}}{2 x^2}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + 3x} - \sqrt{1 - 3x}}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{2 - x} - \sqrt{2 + x}}{x}\]
\[\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - 10}\]
\[\lim_{x \to 0} \frac{a^x + b^x - 2}{x}\]
\[\lim_{x \to \infty} \left( a^{1/x} - 1 \right)x\]
\[\lim_{x \to 0} \frac{e^x - 1 + \sin x}{x}\]
\[\lim_{x \to 0} \frac{e\sin x - 1}{x}\]
\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\]
\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\]
`\lim_{x \to \pi/2} \frac{a^\cot x - a^\cos x}{\cot x - \cos x}`
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
\[\lim_{x \to 0} \frac{e^{x + 2} - e^2}{x}\]
`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`
\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\]
\[\lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x}\]
\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`
\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]
\[\lim_{x \to 0} \frac{x\left( e^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
Write the value of \[\lim_{n \to \infty} \frac{n! + \left( n + 1 \right)!}{\left( n + 1 \right)! + \left( n + 2 \right)!} .\]
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 ______.
