Question

solve using L hospital rule​

Answer 1

Answer:

1.2/3

2.0

3.1/3

4.-2/3

this is the answer

Answer 2

Answer:

The value of $k$ is $\dfrac{2}{3}$

Step-by-step explanation:

L'hospital rule says that, if we get indeterminate form when applying the  limit, like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , we need to differentiate the numerator and denominator and apply the limit.

Here,

$\lim\limits_{x \to 0} \dfrac{\log(3+x)-\log(3-x)}{x}=\dfrac{0}{0}$  , which is an indeterminate form.

So, we can differentiate numerator and denominator separately.

$\dfrac{d}{dx} (\log(3+x)-\log(3-x))=\dfrac{1}{3+x} -\left(\dfrac{1}{3-x} (-1)\right)=\dfrac{1}{3+x}+\dfrac{1}{3-x}$

$\dfrac{d}{dx}(x)=1$

Using L'hospital rule

$\lim\limits_{x \to 0}\dfrac{\frac{d}{dx}(\log(3+x)-\log(3-x) )}{\frac{d}{dx}x } =\lim\limits_{x \to 0} \dfrac{\frac{1}{3+x}+\frac{1}{3-x} }{1} =\dfrac{1}{3} +\dfrac{1}{3}=\dfrac{2}{3}$

So value of $k$ is $\dfrac{2}{3}$