# Evaluate : $$\displaystyle{\lim_{x \to 0}}$$ $$xln(1+2tanx)\over 1-cosx$$

## Solution :

$$\displaystyle{\lim_{x \to 0}}$$ $$xln(1+2tanx)\over 1-cosx$$

= $$\displaystyle{\lim_{x \to 0}}$$ $$xln(1+2tanx)\over {1-cosx\over x^2}.x^2$$.$$2tanx\over 2tanx$$

= 4

### Similar Questions

Evaluate : $$\displaystyle{\lim_{x \to 0}}$$ $$x^3 cotx\over {1-cosx}$$

Evaluate : $$\displaystyle{\lim_{x \to \infty}}$$ $$({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}$$

Evaluate the limit : $$\displaystyle{\lim_{x \to \infty}}$$ $$x^2 + x + 1\over {3x^2 + 2x – 5}$$

Evaluate : $$\displaystyle{\lim_{x \to 0}}$$ $$(2+x)sin(2+x)-2sin2\over x$$

If $$\displaystyle{\lim_{x \to \infty}}$$($${x^3+1\over x^2+1}-(ax+b)$$) = 2, then find the value of a and b.