To find: \(\lim\limits_{\text x \to0}\cfrac{2\text x-sin\,\text x}{tan\,\text x+\text x}\)
\(\lim\limits_{\text x \to0}\cfrac{2\text x-sin\,\text x}{tan\,\text x+\text x}\)
Dividing numerator and denominator by x:
We know,
Therefore,
Hence, the value of \(\lim\limits_{\text x \to0}\cfrac{2\text x-sin\,\text x}{tan\,\text x+\text x}\) = \(\cfrac12\)