How do you find the limit $x^2/sqrt(2x+1)-1$ as $x->0$ ?

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Answer 1 · 2017-01-02T16:33:54

As asked, at $x=0$ , the denominator is not $0$ . So use substitution.

For $lim_(xrarr0)x^2/(sqrt(2x+1)-1)$ see below.

As asked

For $x$ close to $0$ , the numerator, $x^2$ is also close to zero.

The denominator is close to $sqrt(2(0)+1) = sqrt1 = 1$

So the quotient is close to $0/1$ which is $0$ .

If a different quotient was intended

If the question is missing parentheses and should be x^2/(sqrt(2x+1)-1), then use

$x^2/(sqrt(2x+1)-1) = x^2/((sqrt(2x+1)-1)) * ((sqrt(2x+1)+1))/((sqrt(2x+1)+1))$

$= (x^2(sqrt(2x+1)+1))/((2x+1)-1)$

$= (x^2(sqrt(2x+1)+1))/(2x)$

$= (x(sqrt(2x+1)+1))/2$ .

So, the limit is $(0(sqrt(2(0)+1)+1))/2 = 0$

How do you find the limit x^2/sqrt(2x+1)-1x^2/sqrt(2x+1)-1 as x->0x->0?