Prove that the following limit using the epsilon-delta definition:
$$ \lim_{x \to \infty} \frac{x+\sqrt{x}}{x+1} = 1 $$
By definition, with $ \epsilon > 0 $, there must be a $ N > x $ so
$$ N > x \implies \lvert \frac{x+\sqrt{x}}{x+1} – 1 \rvert < \epsilon $$
I know how to solve more simple problems of this instance, but this function is giving me a hard time because of the separate radical in the numerator, so I'm unable to isolate the $ x $ in one side.
Best Answer
Well: $$\left \lvert \frac {x + \sqrt x} {x + 1} - 1 \right \rvert = \left \lvert \frac {x + \sqrt x} {x + 1} - \frac {x + 1} {x + 1} \right \rvert = \left \lvert \frac {\sqrt x - 1} {x + 1} \right \rvert < \left \lvert \frac {\sqrt x - 1} {x - 1} \right \rvert = \left \lvert \frac 1 {\sqrt x + 1} \right \rvert$$
(Edit: Inequality added. Valid for $x > 1$.)
Now to find some $N$, depending on $\epsilon$, such that for all $x \ge N$, the above is less than $\epsilon$.