Let $h(x)=\frac{x-2}{\sqrt{x+7}-3}$ when $x
eq 2$.
$h$ is continuous for all $x-7$.
Find $h(2)$.

Question

Let $h(x)=\frac{x-2}{\sqrt{x+7}-3}$ when $x 
eq 2$.
$h$ is continuous for all $x>-7$.
Find $h(2)$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find $h(2)$, we need to evaluate the limit of $h(x)$ as $x$ approaches 2. ‖ ‖ step 2 ⋮ The limit is given by $\lim_{x o 2} \frac{x - 2}{\sqrt{x + 7} - 3}$. ‖ ‖ step 3 ⋮ Direct substitution of $x = 2$ into $h(x)$ results in an indeterminate form $\frac{0}{0}$. ‖ ‖ step 4 ⋮ To resolve this, we can use the Taylor series expansion around $x = 2$. ‖ ‖ step 5 ⋮ The Taylor series expansion provided by asksia-ll calculator is $6 + \frac{x - 2}{6} - \frac{1}{216} (x - 2)^2 + \frac{(x - 2)^3}{3888} - \frac{5 (x - 2)^4}{279936} + \frac{7 (x - 2)^5}{5038848} + O((x - 2)^6)$. ‖ ‖ step 6 ⋮ Evaluating the limit using the Taylor series, we find that $\lim_{x o 2} \frac{x - 2}{\sqrt{x + 7} - 3} = 6$. ‖ ‖ step 7 ⋮ Therefore, $h(2) = 6$. ‖ ∻Answer∻ ⚹ $h(2) = 6$ ⚹ ∻Key Concept∻ ⚹Evaluating limits using Taylor series⚹ ∻Explanation∻ ⚹When direct substitution in a limit results in an indeterminate form, a Taylor series expansion can be used to evaluate the limit. In this case, the expansion around $x = 2$ for the function $h(x)$ gives the value of the limit as $x$ approaches 2.⚹◊What is the limit of $h(x)$ as $x$ approaches 2?⍭ Generate me a similar question◊

Sia