Question-162834

Question Number 162834 by mkam last updated on 01/Jan/22

Answered by abdullahhhhh last updated on 01/Jan/22

\lim_{x \to 0} \frac{x + \tan 2 x}{x - \tan 2 x} / x

\lim_{x \to 0} \frac{1 + 2}{1 - 2} = - 3

Answered by abdullahhhhh last updated on 01/Jan/22

\lim_{x \to e} \frac{sinx - x}{x^{3}} = \frac{\sin (e) - e}{e^{3}}

Answered by abdullahhhhh last updated on 01/Jan/22

\lim_{x \to 0} \frac{e^{x} + e^{- x} - x^{2} - 2}{\sin^{2} x - x^{2}} = (\frac{0}{0}) use lobital

\lim_{x \to 0} \frac{e^{x} - e^{- x} - 2 x}{2 sinx cosx (\sin 2 x) - 2 x} = \frac{0}{0}

\lim_{x \to 0} \frac{e^{x} + e^{- x} - 2}{2 \cos 2 x - 2} = \frac{2 - 2}{2 - 2} = \frac{0}{0}

\lim_{x \to 0} \frac{e^{x} - e^{- x}}{- 4 \sin 2 x} = \frac{1 - 1}{0} = \frac{0}{0}

\lim_{x \to 0} \frac{e^{x} + e^{- x}}{- 8 \cos 2 x} = \frac{2}{- 8} = \frac{- 1}{4}

Answered by abdullahhhhh last updated on 01/Jan/22

\lim_{x \to 0} \frac{{xe}^{x} - \ln (1 + x)}{x^{2}} = \frac{0}{0} lobital

\lim_{x \to 0} \frac{{xe}^{x} + e^{x} - \frac{1}{(x + 1)}}{2 x} = \frac{0}{0}

\lim_{x \to 0} \frac{{xe}^{x} + e^{x} + e^{x} + \frac{1}{{(x + 1)}^{2}}}{2} = \frac{3}{2}

Answered by abdullahhhhh last updated on 01/Jan/22

\lim_{x \to \infty} (\frac{lnx}{x}) = \frac{\infty}{\infty} use lobital

\lim_{x \to \infty} \frac{1}{x} = \frac{1}{\infty} = 0