Advertisements
Advertisements
प्रश्न
Evaluate `lim_(x -> 0) f(x)` where `f(x) = { (|x|/x, x != 0),(0, x = 0):}`
Advertisements
उत्तर
If x < 0, |x| = −x
∴ `lim_("x" → 0^-) f("x") = lim_("x" → 0^-) |"x"|/"x" = lim_("x" → 0^-)( (-"x")/"x") = -1`
And if x > 0, |x| = x
∴ `lim_("x" → 0^+) f("x") = lim_("x" → 0^+) |"x"|/"x" = lim_("x" → 0^+) ( "x"/"x") = 1`
∴ `lim_("x" → 0^-) f("x") ≠ lim_("x" → 0^+) f("x")`
Hence, the equation does not exist at x = 0.
APPEARS IN
संबंधित प्रश्न
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)`
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?
\[\lim_{x \to 0} \frac{2x}{\sqrt{a + x} - \sqrt{a - x}}\]
\[\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}\]
\[\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}\]
\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\]
\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\]
\[\lim_{x \to 4} \frac{2 - \sqrt{x}}{4 - x}\]
\[\lim_{x \to a} \frac{x - a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}\]
\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\]
\[\lim_{x \to 0} \frac{a^{mx} - 1}{b^{nx} - 1}, n \neq 0\]
\[\lim_{x \to 0} \frac{9^x - 2 . 6^x + 4^x}{x^2}\]
\[\lim_{x \to 2} \frac{x - 2}{\log_a \left( x - 1 \right)}\]
\[\lim_{x \to 0} \frac{e\sin x - 1}{x}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log \left( a - x \right)}{x}\]
\[\lim_{x \to 0} \frac{x\left( 2^x - 1 \right)}{1 - \cos x}\]
`\lim_{x \to \pi/2} \frac{a^\cot x - a^\cos x}{\cot x - \cos x}`
\[\lim_{x \to 5} \frac{e^x - e^5}{x - 5}\]
\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\]
\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`
\[\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}\]
\[\lim_{x \to 0} \frac{x\left( e^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
Evaluate: `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 2))`
