limit : calculus

[abhishekrukhaiyar]

A. solve the limit without using L'Hospital or expansion.
1. lt(x-->inf) [x - {x^2 * ln(i + 1/x)}]
2. lt(x-->0) {e^x - e^(-x) - 2x}/{x - sin x}
B. Evaluate lt(x-->0) [x*ln(x)]

 

[galactus]

Are these your limits?:

\(\displaystyle \L\\\lim_{x\to\infty}\left[x-x^{2}ln(1+\frac{1}{x})\right]\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{e^{x}-e^{-x}-2x}{x-sin(x)}\)

\(\displaystyle \L\\\lim_{x\to\0}xln(x)\)