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Mathematics in the high school

Mathematics in the high school-example-1

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

User Mkurz
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