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

The limit as x approaches 1 of sqrt(6x^2+1) is square root of 7

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limit as x approaches 1 of sqrt(6x^2+1)

x \to 1 lim (6 x^{2} + 1)

7

Decimal Notation

2.64575 \dots

Solution steps

x \to 1 lim (6 x^{2} + 1)

Plug in the value

x = 1

= 6 \cdot 1^{2} + 1

Simplify

= 7

\int \frac{1}{4 + v ^{2}} d v

f (x) = \frac{1 - x ^{2}}{x}

y = 10 csc (5 x^{5} - 4 x + 4)

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

What is the limit as x approaches 1 of sqrt(6x^2+1) ?
The limit as x approaches 1 of sqrt(6x^2+1) is square root of 7