Find the value of the limit $\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } + x } - x \right)$
A)
$- \frac { 1 } { 2 }$
B) $\sqrt { 2 }$
C)
$- \sqrt { 2 }$
D) $\frac { 1 } { 4 }$
E)
$\infty$
F) $- \frac { 1 } { 4 }$
G)0
H)$\frac { 1 } { 2 }$

Question

Find the value of the limit $\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } + x } - x \right)$ 
A)
$- \frac { 1 } { 2 }$
B) $\sqrt { 2 }$ 
C)
$- \sqrt { 2 }$ 
D) $\frac { 1 } { 4 }$
E)
$\infty$ 
F) $- \frac { 1 } { 4 }$
G)0
H)$\frac { 1 } { 2 }$

Quizplus · Accepted Answer

To solve this limit, we can multiply and divide by the conjugate of the expression inside the limit: $\sqrt{x^2 + x} + x$. This gives us:$$\lim _ { x \rightarrow \infty } \left( \frac { (\sqrt { x ^ { 2 } + x } - x)(\sqrt { x ^ { 2 } + x } + x) }{\sqrt { x ^ { 2 } + x } + x} \right) = \lim _ { x \rightarrow \infty } \left( \frac { x^2 + x - x^2 }{\sqrt { x ^ { 2 } + x } + x} \right) = \lim _ { x \rightarrow \infty } \left( \frac { x }{\sqrt { x ^ { 2 } + x } + x} \right)$$Simplifying further by dividing numerator and denominator by $x$, we get:$$\lim _ { x \rightarrow \infty } \left( \frac { 1 }{\sqrt { 1 + \frac{1}{x} } + 1} \right) = \frac { 1 }{\sqrt { 1 + 0 } + 1} = \frac { 1 }{2}$$Therefore, the correct answer is $\frac { 1 } { 2 }$.

Find the Value of the Limit lim⁡x→∞(x2+x−x)\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } + x } - x \right)limx→∞​(x2+x​−x)

Find the Value of the Limit $\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } + x } - x \right)$