lim sqrt(1-t) + sqrt(1+t) t 0 lim (2t2 + 3t + 1)/(t2 - 1) t infinity

Question

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to your limit problems.

1.) To evaluate the limit, substitute $t=0$ into the expression.

Substitute $t=0$ into the function.
$\lim_{t \to 0} (\sqrt{1-t} + \sqrt{1+t}) = \sqrt{1-0} + \sqrt{1+0}$
Simplify the expression.
$= \sqrt{1} + \sqrt{1} = 1 + 1 = 2$
The limit is:
$\boxed{2}$
2.) To evaluate the limit as $t \to \infty$ , divide every term in the numerator and denominator by the highest power of $t$ in the denominator, which is $t^2$ .
Divide all terms by $t^2$ .
$\lim_{t \to \infty} \frac{2t^2 + 3t + 1}{t^2 - 1} = \lim_{t \to \infty} \frac{\frac{2t^2}{t^2} + \frac{3t}{t^2} + \frac{1}{t^2}}{\frac{t^2}{t^2} - \frac{1}{t^2}}$
Simplify the terms.
$= \lim_{t \to \infty} \frac{2 + \frac{3}{t} + \frac{1}{t^2}}{1 - \frac{1}{t^2}}$
Apply the limit property $\lim_{t \to \infty} \frac{c}{t^n} = 0$ for $n > 0$ .
$= \frac{2 + 0 + 0}{1 - 0}$
Simplify the result.
$= \frac{2}{1} = 2$

Answer