Question

Mathematics in the high school

Answer 1

The limit
`\(\lim_{{x \to 0}} \left[ (1+x^2) \cdot \frac{{2x+1}}{{x \sin(x)}} \right]\)` is evaluated using L'Hôpital's Rule, simplifying to
`\(\frac{{2}}{{1}}\)`, resulting in a limit of 2 as x approaches 0.

To evaluate the limit
`\(\lim_{{x \to 0}} \left[ (1+x^2) \cdot \frac{{2x+1}}{{x \sin(x)}} \right]\)`, we'll apply algebraic simplification and use known limits.

1. **Factorization:**

`\[\lim_{{x \to 0}} \left[ (1+x^2) \cdot \frac{{2x+1}}{{x \sin(x)}} \right] = \lim_{{x \to 0}} \frac{{(1+x^2)(2x+1)}}{{x \sin(x)}}\]`

2. **Apply L'Hôpital's Rule:**

`\[ \lim_{{x \to 0}} \frac{{(1+x^2)(2x+1)}}{{x \sin(x)}} = \frac{{(2x)(1+x^2) + (2)(2x+1)(x)}}{{\sin(x) + x \cos(x)}}\]`

Simplify further:

`\[ = \frac{{2x + 2x^3 + 4x + 2}}{{\sin(x) + x \cos(x)}} \]`

`\[= \frac{{2x^3 + 6x + 2}}{{\sin(x) + x \cos(x)}} \]`

3. **Evaluate the Limit:**

Now, substitute x = 0:

`\[\lim_{{x \to 0}} \frac{{2x^3 + 6x + 2}}{{\sin(x) + x \cos(x)}} = \frac{{2}}{{1}} = 2\]`

Therefore,
`\(\lim_{{x \to 0}} \left[ (1+x^2) \cdot \frac{{2x+1}}{{x \sin(x)}} \right] = 2\).`