Advertisements
Advertisements
प्रश्न
Evaluate the following limits: `lim_(x -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
Advertisements
उत्तर
`lim_(x -> "a")((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")`
Put z + 2 = y and a + 2 = b
As z → a, z + 2 → a + 2
i.e. y → b
∴ `lim_(z -> "a") ((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")`
= `lim_(y -> "b") (y^(3/2) - "b"^(3/2))/((y - 2) - ("b" - 2))`
= `lim_(y -> "b") (y^(3/2) - "b"^(3/2))/(y - "b")`
= `3/2*"b"^(1/2) ...[because lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`
= `3/2("a" + 2)^(1/2)` ...[∵ b = a + 2]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}\]
\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
Evaluate the following limit:
\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\]
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number.
\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1
Evaluate the following limit:
\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
Write the value of \[\lim_{x \to \infty} \frac{\sin x}{x} .\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
Write the value of \[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
The value of \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is
Evaluate the following limits: if `lim_(x -> 1)[(x^4 - 1)/(x - 1)] = lim_(x -> "a") [(x^3 - "a"^3)/(x - "a")]`, find all the value of a.
Evaluate the following limits: `lim_(y -> 1) [(2y - 2)/(root(3)(7 + y) - 2)]`
Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`
Evaluate the following limit :
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
