Advertisements
Advertisements
प्रश्न
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
Advertisements
उत्तर १
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
Let x =\[-\] m When n → – ∞, then m → ∞.
\[ = \lim_{m \to \infty} \left[ \left( \sqrt{4 m^2 = 7m} - 2m \right) \times \frac{\left( \sqrt{4 m^2 + 7m} + 2m \right)}{\left( \sqrt{4 m^2 + 7m} + 2m \right)} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{\left( 4 m^2 + 7m \right) - \left( 2m \right)^2}{\sqrt{4 m^2 + 7m} + 2m} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{4 m^2 + 7m - 4 m^2}{\sqrt{4 m^2 + 7m} + 2m} \right]\]
Dividing the numerator and the denominator by m:
\[\lim_{m \to \infty} \left[ \frac{7}{\sqrt{\frac{4 m^2 + 7m}{m^2}} + \frac{2m}{m}} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{7}{\sqrt{\frac{4 m^2}{m^2} + \frac{7m}{m^2}} + 2} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{7}{\sqrt{4 + \frac{7}{m}} + 2} \right]\]
\[\text{ As } m \to \infty , \frac{1}{m} \to 0\]
\[ = \frac{7}{\sqrt{4} + 2}\]
\[ = \frac{7}{4}\]
उत्तर २
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
Let x =\[-\] m When n → – ∞, then m → ∞.
\[ = \lim_{m \to \infty} \left[ \left( \sqrt{4 m^2 = 7m} - 2m \right) \times \frac{\left( \sqrt{4 m^2 + 7m} + 2m \right)}{\left( \sqrt{4 m^2 + 7m} + 2m \right)} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{\left( 4 m^2 + 7m \right) - \left( 2m \right)^2}{\sqrt{4 m^2 + 7m} + 2m} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{4 m^2 + 7m - 4 m^2}{\sqrt{4 m^2 + 7m} + 2m} \right]\]
Dividing the numerator and the denominator by m:
\[\lim_{m \to \infty} \left[ \frac{7}{\sqrt{\frac{4 m^2 + 7m}{m^2}} + \frac{2m}{m}} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{7}{\sqrt{\frac{4 m^2}{m^2} + \frac{7m}{m^2}} + 2} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{7}{\sqrt{4 + \frac{7}{m}} + 2} \right]\]
\[\text{ As } m \to \infty , \frac{1}{m} \to 0\]
\[ = \frac{7}{\sqrt{4} + 2}\]
\[ = \frac{7}{4}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]
\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\]
\[\lim_{x \to \sqrt{2}} \frac{x^2 - 2}{x^2 + \sqrt{2}x - 4}\]
\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\]
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = \lim_{x \to 5} \left( 4 + x \right),\] find all possible values of a.
\[\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}\]
\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\]
\[\lim_{x \to 0} \frac{\sin 3x}{5x}\]
\[\lim_{x \to 0} \frac{3 \sin 2x + 2x}{3x + 2 \tan 3x}\]
\[\lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x}\]
\[\lim_{x \to 0} \frac{2 \sin x^\circ - \sin 2 x^\circ}{x^3}\]
Evaluate the following limits:
\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\]
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
\[\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3}\]
Evaluate the following limit:
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\]
Evaluate the following limit:
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\]
\[\lim_{x \to 1} \frac{1 + \cos \pi x}{\left( 1 - x \right)^2}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\cos x - \sin x}{\left( \frac{\pi}{4} - x \right) \left( \cos x + \sin x \right)}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{{cosec}^2 x - 2}{\cot x - 1}\]
\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]
\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
Write the value of \[\lim_{x \to \infty} \frac{\sin x}{x} .\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 - n^2} + \frac{2}{1 - n^2} + . . . + \frac{n}{1 - n^2} \right\}\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to
\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to
\[\lim_{x \to \infty} \frac{\left| x \right|}{x}\] is equal to
Number of values of x where the function
f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3"," π):}`
is discontinuous, is ______.
Evaluate the following limits: `lim_(x ->3) [sqrt(x + 6)/x]`
Evaluate the following limit :
`lim_(x->3)[sqrt(x+6)/x]`
Evaluate the following limit.
`lim_(x->3)[sqrt(x + 6)/x]`
Evaluate the Following limit:
`lim_(x->7)[((root(3)(x)-root(3)(7))(root(3)(x)+root(3)(7)))/(x-7)]`
Evaluate the following limit:
`\underset{x->3}{lim}[sqrt(x +6)/(x)]`
Evaluate the Following limit:
`lim_(x->7)[[(root[3][x] - root[3][7])(root[3][x] + root[3][7])] / (x - 7)]`
