Question

Math

Posted 5 months ago

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

Sign in to unlock the answer

Answer from Sia

Posted 5 months ago

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 $\lim_{x \to 2} \frac{x - 2}{\sqrt{x + 7} - 3}$

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 $\lim_{x \to 2} \frac{x - 2}{\sqrt{x + 7} - 3} = 6$

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.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Upgrade to Asksia Pro

Join a AskSia's Pro Plan, and get 24/7 AI tutoring for your reviews, assignments, quizzes and exam preps.

Unlimited chat query usages

Strong algorithms that better know you

Early access to new release features

Study Other Question