Advertisements
Advertisements
Question
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
Advertisements
Solution
\[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\text{ LHS }: \]
\[ \lim_{x \to \infty} \left( \left( \sqrt{x^2 + x + 1} - x \right) \right)\]
\[\text{ Rationalising the numerator }: \]
\[ \lim_{x \to \infty} \left[ \frac{\left( \sqrt{x^2 + x + 1} - x \right) \left( \sqrt{x^2 + x + 1} + x \right)}{\left( \sqrt{x^2 + x + 1} + x \right)} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{\left( x^2 + x + 1 \right) - x^2}{\left( \sqrt{x^2 + x + 1} + x \right)} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{x + 1}{\left( \sqrt{x^2 + x + 1} + x \right)} \right]\]
\[\text{ Dividing the numerator and the denominator by x }: \]
\[ \lim_{x \to \infty} \left[ \frac{1 + \frac{1}{x}}{\frac{\sqrt{x^2 + x + 1}}{x} + 1} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{1 + \frac{1}{x}}{\sqrt{\frac{x^2 + x + 1}{x^2}} + 1} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + 1} \right]\]
\[\text{ When x } \to \infty , \text{ then } \frac{1}{x} \to 0 . \]
\[\frac{1}{\sqrt{1} + 1}\]
\[ = \frac{1}{2}\]
\[RHS: \]
\[ \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right) \left[ \text{ from } \infty - \infty \right]\]
Rationalising the numerator:
\[\lim_{x \to \infty} \left[ \frac{\left( \sqrt{x^2 + 1} - x \right) \left( \sqrt{x^2 + 1} + x \right)}{\left( \sqrt{x^2 + 1} + x \right)} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{x^2 + 1 - x^2}{\left( \sqrt{x^2 + 1} + x \right)} \right]\]
\[ = \frac{1}{\infty}\]
\[ = 0\]
\[ \therefore \lim_{x \to \infty} \left[ \sqrt{x^2 + x + 1} - x \right] \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
APPEARS IN
RELATED QUESTIONS
Suppose f(x) = `{(a+bx, x < 1),(4, x = 1),(b-ax, x > 1):}` and if `lim_(x -> 1) f(x) = f(1)` what are possible values of a and b?
\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\]
\[\lim_{x \to - 2} \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2}\]
\[\lim_{x \to a} \frac{x^{2/7} - a^{2/7}}{x - a}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\]
\[\lim_{n \to \infty} \frac{n^2}{1 + 2 + 3 + . . . + n}\]
\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\]
\[\lim_{x \to 0} \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to
\[\lim 2_{h \to 0} \left\{ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h \left( \sqrt{3} \cos h - \sin h \right)} \right\}\]
If \[f\left( x \right) = \left\{ \begin{array}{l}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0\end{array}, \right.\] then \[\lim_{x \to 0} f\left( x \right)\] equals
If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]
The value of \[\lim_{n \to \infty} \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!}\] is
The value of \[\lim_{n \to \infty} \left\{ \frac{1 + 2 + 3 + . . . + n}{n + 2} - \frac{n}{2} \right\}\]
Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
Which of the following function is not continuous at x = 0?
Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`
`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->7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
Evaluate the Following limit:
`lim_ (x -> 3) [sqrt (x + 6)/ x]`
