Question

question 10 (1 point)
the limit
lim _ { x
ightarrow 0 } \frac { int _ { x } ^ { 0 } ln left( 2 + t ^ { 2 }
ight) d t } { x }
is equal to
0
-∞

• ln(2)

∞
ln(2)

Explanation:

Step1: Check indeterminate form

As $x \to 0$, numerator $\int_{x}^{0}\ln(2+t^2)dt = -\int_{0}^{x}\ln(2+t^2)dt \to 0$, denominator $x \to 0$. So it's $\frac{0}{0}$ form.

Step2: Apply L'Hospital's Rule

Differentiate numerator and denominator:
Numerator derivative: $\frac{d}{dx}\int_{x}^{0}\ln(2+t^2)dt = -\ln(2+x^2)$
Denominator derivative: $\frac{d}{dx}(x) = 1$
Limit becomes $\lim_{x \to 0} -\ln(2+x^2)$

Step3: Substitute $x=0$

$-\ln(2+0^2) = -\ln2$

Answer:

$-\ln(2)$