Advertisements
Advertisements
Question
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
Advertisements
Solution 1
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
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:
\[\lim_{m \to \infty} \left[ \frac{7}{\sqrt{\frac{4 m^2 + 7m}{m^2}} + \frac{2m}{m}} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{7}{\sqrt{\frac{4 m^2}{m^2} + \frac{7m}{m^2}} + 2} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{7}{\sqrt{4 + \frac{7}{m}} + 2} \right]\]
\[\text{ As } m \to \infty , \frac{1}{m} \to 0\]
\[ = \frac{7}{\sqrt{4} + 2}\]
\[ = \frac{7}{4}\]
Solution 2
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
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
RELATED QUESTIONS
Find `lim_(x -> 5) f(x)`, where f(x) = |x| - 5
Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.
\[\lim_{x \to 3} \frac{\sqrt{2x + 3}}{x + 3}\]
\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]
\[\lim_{x \to - 1} \frac{x^3 - 3x + 1}{x - 1}\]
\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\]
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 3} \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3}\]
\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]
\[\lim_{x \to 1} \left\{ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3 x^2 + 2x} \right\}\]
\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\]
\[\lim_{x \to 0} \frac{\sin x^n}{x^n}\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{3 \sin^2 x - 2 \sin x^2}{3 x^2}\]
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
\[\lim_{x \to 0} \frac{\sin 3x + 7x}{4x + \sin 2x}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\]
\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\]
\[\lim_{n \to \infty} 2^{n - 1} \sin \left( \frac{a}{2^n} \right)\]
\[\lim_{x \to 1} \left( 1 - x \right) \tan \left( \frac{\pi x}{2} \right)\]
Write the value of \[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
\[\lim 2_{h \to 0} \left\{ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h \left( \sqrt{3} \cos h - \sin h \right)} \right\}\]
The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\] is
\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to
\[\lim_{x \to 0} \frac{\left| \sin x \right|}{x}\]
Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`
if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.
`1/(ax^2 + bx + c)`
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->5)[(x^3-125)/(x^5-3125)]`
Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`
Evaluate the following limit:
`\underset{x->5}{lim}[(x^3 - 125)/(x^5 - 3125)]`
