What is the limit as x approaches 4 of sqrt(x^2-9) ?

The limit as x approaches 4 of sqrt(x^2-9) is square root of 7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

limit as x approaches 4 of sqrt(x^2-9)

x \to 4 lim (x^{2} - 9)

7

Decimal

2.64575 \dots

Solution steps

x \to 4 lim (x^{2} - 9)

Plug in the value

x = 4

= 4^{2} - 9

Simplify

= 7

\int - 4 x^{\frac{3}{2}} + 3 x^{\frac{2}{3}} - 4 x^{2} d x

d er i v a t i v e y = e^{- 6 x^{5}}

\int cot^{6} (x) csc^{4} (x) d x

in v erse l a pl a ce \frac{1}{( s + 1 ) ( s + 2 )}

x^{2} y^{^{''}} - 2 x y^{^{'}} + (2 + x^{2}) y = 0

What is the limit as x approaches 4 of sqrt(x^2-9) ?
The limit as x approaches 4 of sqrt(x^2-9) is square root of 7