Find $ \lim_{t\to 0}\frac{\sqrt{1+t}-\sqrt{1-t}}{t}. $
I can't think of how to start this or what to do at all. Anything I try just doesn't change the function.
Find $ \lim_{t\to 0}\frac{\sqrt{1+t}-\sqrt{1-t}}{t}. $
I can't think of how to start this or what to do at all. Anything I try just doesn't change the function.
HINT $\ $ Use the same method in your prior question, i.e. rationalize the numerator by multiplying both the numerator and denominator by the numerator's conjugate $\rm\:\sqrt{1+t}+\sqrt{1-t}\:.$ Then the numerator becomes $\rm\:(1+t)-(1-t) = 2\:t,\:$ which cancels with the denominator $\rm\:t\:,\:$ so $\rm\:\ldots$
More generally, using the same notation and method as in your prior question, if $\rm\:f_0 = g_0\:$ then
$\rm \lim_{x\:\to\: 0}\ \dfrac{\sqrt{f(x)}-\sqrt{g(x)}}{x}\ = \ \lim_{x\:\to\: 0}\ \dfrac{f(x)-g(x)}{x\ (\sqrt{f(x)}+\sqrt{g(x)})}\ =\ \dfrac{f_1-g_1}{\sqrt{f_0}+\sqrt{g_0}}$
In your case $\rm\: f_0 = 1 = g_0,\ \ f_1 = 1,\ g_1 = -1\:,\ $ so the limit $\: =\: (1- (-1))/(1+1)\: =\: 1\:.$
Note again, as in your prior questions, rationalizing the numerator permits us to cancel the common factor at the heart of the indeterminacy - thus removing the apparent singularity.
$\frac{\sqrt{1+t}-\sqrt{1-t}}{t}$
$=\frac{\sqrt{1+t}-\sqrt{1-t}}{t}\cdot\frac{(\sqrt{1+t}+\sqrt{1-t})}{(\sqrt{1+t}+\sqrt{1-t})}$
Note $(a-b)(a+b)=a^2-b^2$, so this is
$\frac{(\sqrt{1+t})^2-(\sqrt{1-t})^2}{t(\sqrt{1+t}+\sqrt{1-t})}$
$=\frac{(1+t)-(1-t)}{t(\sqrt{1+t}+\sqrt{1-t})}$
The top is $2t$, so this is
$\frac{2}{\sqrt{1+t}+\sqrt{1-t}}.$
To find this as $t\to0$ just plug in $t=0$, which gives
$\frac{2}{\sqrt{1}+\sqrt{1}}=1.$
This is probably not what you are expected to do (you probably are not supposed to know/use Taylor series at this point), but for the sake of information: this kind of limits are more easily solved with Taylor expansions. Knowing that around zero $\sqrt{1+x} = 1 + \frac{x}{2} + O(x^2)$, the result comes immediately.