# How do you find lim t(sqrt(t+1)-sqrtt) as t->oo?

Jan 7, 2018

${\lim}_{t \to \infty} t \left(\sqrt{t + 1} - \sqrt{t}\right) = \infty$

#### Explanation:

${\lim}_{t \to \infty} t \left(\sqrt{t + 1} - \sqrt{t}\right)$

$= {\lim}_{t \to \infty} \frac{t \left(\sqrt{t + 1} - \sqrt{t}\right) \left(\sqrt{t + 1} + \sqrt{t}\right)}{\sqrt{t + 1} + \sqrt{t}}$

$= {\lim}_{t \to \infty} \frac{t \left(\left(t + 1\right) - t\right)}{\sqrt{t + 1} + \sqrt{t}}$

$= {\lim}_{t \to \infty} \frac{t}{\sqrt{t + 1} + \sqrt{t}}$

$= {\lim}_{t \to \infty} \sqrt{t} \cdot \frac{\sqrt{t}}{\sqrt{t + 1} + \sqrt{t}}$

$= {\lim}_{t \to \infty} \sqrt{t} \cdot \frac{1}{\sqrt{1 + \frac{1}{t}} + 1}$

$= {\lim}_{t \to \infty} \frac{1}{2} \sqrt{t}$

$= \infty$

Jan 7, 2018

$+ \infty$

#### Explanation:

I'll replace $t$ with $x$

${\lim}_{x \rightarrow + \infty} x \cdot \left(\sqrt{x + 1} - \sqrt{x}\right)$ $=$

${\lim}_{x \rightarrow + \infty} x \cdot \frac{\left(\sqrt{x + 1} - \sqrt{x}\right) \left(\sqrt{x + 1} + \sqrt{x}\right)}{\sqrt{x + 1} + \sqrt{x}}$ $=$

${\lim}_{x \rightarrow + \infty} x \cdot \frac{{\sqrt{x + 1}}^{2} - {\sqrt{x}}^{2}}{\sqrt{x + 1} + \sqrt{x}}$ $=$

${\lim}_{x \rightarrow + \infty} x \cdot \frac{\cancel{x} + 1 - \cancel{x}}{\sqrt{x + 1} + \sqrt{x}}$ $=$

${\lim}_{x \rightarrow + \infty} \frac{x}{\sqrt{x + 1} + \sqrt{x}}$ $=$

${\lim}_{x \rightarrow + \infty} \frac{x}{\sqrt{{x}^{2} \left(\frac{1}{x} + \frac{1}{x} ^ 2\right)} + \sqrt{{x}^{2} \cdot \left(\frac{1}{x}\right)}}$ $=$

${\lim}_{x \rightarrow + \infty} \frac{x}{| x | \sqrt{\frac{1}{x} + \frac{1}{x} ^ 2} + | x | \sqrt{\frac{1}{x}}}$

$x \to + \infty$ , $x > 0$

$=$ ${\lim}_{x \rightarrow + \infty} \frac{x}{x \sqrt{\frac{1}{x} + \frac{1}{x} ^ 2} + x \sqrt{\frac{1}{x}}}$ $=$

lim_(xrarr+oo)cancel(x)/(cancel(x)(sqrt(1/x+1/x^2)+sqrt(1/x)) $=$

${\lim}_{x \rightarrow + \infty} \frac{1}{\sqrt{\frac{1}{x} + \frac{1}{x} ^ 2} + \sqrt{\frac{1}{x}}}$ $=$ $+ \infty$

because ${\lim}_{x \rightarrow + \infty} x = + \infty$ so ${\lim}_{x \rightarrow + \infty} \frac{1}{x} = 0$

and ${\lim}_{x \rightarrow + \infty} \sqrt{\frac{1}{x}} = 0$