Lim x- - - x 2 -ax -x - 1 2 then a-

The correct option is C 1 lim_x→ ∞√(x^2 ax) x=lim_x→ ∞(√(()x^2 ax) x)(x^2 ax)+x√(x^2 ax)+x =lim_x→ ∞x^2 ax x^2x[√(1a/x)+1]=lim_x→ ∞ axx[√(1 a ...

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Functions

lim x→ -∞ √ x...

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$lim x \to - \infty (\sqrt{x^{2} - a x} - x) = \frac{1}{2}$ then $a =$

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lim x \to \infty (\frac{x^{2} - 1}{x + 1} - a x - b) = 2

Evaluate the following one sided limits:

$(i) {lim}_{x \to 2^{+}} \frac{x - 3}{x^{2} - 4}$

$(i i) {lim}_{x \to 2^{-}} \frac{x - 3}{x^{2} - 4}$

$(i i i) {lim}_{x \to 0^{+}} \frac{1}{3 x}$

$(i v) {lim}_{x \to 8^{+}} \frac{2 x}{x + 8}$

$(v) {lim}_{x \to 0^{+}} \frac{2}{x^{\frac{1}{5}}}$

$(v i) {lim}_{x \to \frac{π^{-}}{2}} t a n x$

$(v i i) {lim}_{x \to \frac{π}{2^{+}}} s e c x$

$(v i i i) {lim}_{x \to 0^{-}} \frac{x^{2} - 3 x + 2}{x^{3} - 2 x^{2}}$

$(i x) {lim}_{x \to - 2^{+}} \frac{x^{2} - 1}{2 x + 4}$

$(x) {lim}_{x \to 0^{+}} (2 - c o t x)$

$(x i) {lim}_{x \to 0^{-}} 1 + c o s e c x$

lim x \to \infty (\frac{x^{3} + 1}{x^{2} + 1} - (a x + b)) = 2

lim x \to \infty \frac{x^{3} + 1}{x^{2} + 1} - (a x + b) = 2

lim x \to \infty (\sqrt{x^{2} - x + 1} - a x - b) = 0,

k \geq 2,

lim n \to \infty {sec}^{2 n} (k! π b)

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