What is the limit as x approaches 1 of sqrt(4x^2+2) ?

The limit as x approaches 1 of sqrt(4x^2+2) is square root of 6

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limit as x approaches 1 of sqrt(4x^2+2)

x \to 1 lim (4 x^{2} + 2)

6

Decimal

2.44948 \dots

Solution steps

x \to 1 lim (4 x^{2} + 2)

Plug in the value

x = 1

= 4 \cdot 1^{2} + 2

Simplify

= 6

\frac{4}{x} - 1

\int x e^{\frac{- x ^{2}}{2}} d x

\int_{- 2}^{6} 12 + 4 x - x^{2} d x

\int t^{5} cos (2 t^{6}) d t

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{x}

What is the limit as x approaches 1 of sqrt(4x^2+2) ?
The limit as x approaches 1 of sqrt(4x^2+2) is square root of 6