What is the limit as x approaches 16 of sqrt(x^2-9) ?

The limit as x approaches 16 of sqrt(x^2-9) is square root of 247

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches 16 of sqrt(x^2-9)

x \to 16 lim (x^{2} - 9)

247

Decimal Notation

15.71623 \dots

Solution steps

x \to 16 lim (x^{2} - 9)

Plug in the value

x = 16

= 1 6^{2} - 9

Simplify

= 247

\int \frac{2 x}{( x ^{2} + 1 ) ^{3}} d x

\int tan^{5} (1) d x

\frac{\partial}{\partial y} (ln (3 x + 2 y))

x \to 0 + lim ((x + 1)^{l n (x)})

\frac{d}{d x} (ln (e^{4 x + 1}))

What is the limit as x approaches 16 of sqrt(x^2-9) ?
The limit as x approaches 16 of sqrt(x^2-9) is square root of 247