Lim x- 2-x2-3x2-4-2-34-4-50-But the answer is -Where I am wrong- Please clarity the concept-

lim _x→ 2^ x^2 3x^2 4 lim _x→ 2 nx^2 3x^2 4 lim _n→ 0(2 n)^2 3(2 n)^2 4=2^3 32^2 4=10=∞ ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Theorems for Continuity

lim x→ 2-x2-3...

Question

$lim x \to 2^{-} \frac{x^{2} - 3}{x^{2} - 4} = \frac{- 2 - 3}{4 - 4} = \frac{- 5}{0} = - \infty$
But the answer is $+ \infty$
Where I am wrong. Please clarity the concept.

Open in App

Solution

Suggest Corrections

Similar questions

Evaluate the following one sided limits:

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

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

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

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

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

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

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

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

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

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

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

This is a silly question but due to tension I am not able to figure this out.

(-1-2)^2 according to the formula should be =

1 + 4 - 4 = 1

But the answer is 9.

Where am I wrong?

Find the value of $lim x \to \infty \frac{8 - \frac{3}{x} + \frac{5}{x^{2}}}{4 + \frac{2}{x} + \frac{12}{x^{2}}}$

lim x \to 1 [\frac{3}{x^{2} + x - 2} - \frac{4}{x^{2} + 2 x - 3}]

lim x \to \infty \frac{1}{1 + x^{2}} + \frac{2}{4 + x^{2}} + \frac{3}{9 + x^{2}} + \dots + \frac{x}{x^{2} + x^{2}}

Join BYJU'S Learning Program