# Limit Questions

Solution : $$\displaystyle{\lim_{x \to \infty}}$$$$x^2 + x + 1\over {3x^2 + 2x – 5}$$ It is ($$\infty\over \infty$$ form),   Put x = $$1\over y$$ = $$\displaystyle{\lim_{y \to 0}}$$ $$1 + y + y^2\over {3 + 2y – 5y^2}$$ = $$1\over 3$$ Similar Questions Evaluate : $$\displaystyle{\lim_{x \to 0}}$$ $$x^3 cotx\over {1-cosx}$$ Evaluate : $$\displaystyle{\lim_{x \to … ## Evaluate : \(\displaystyle{\lim_{x \to 0}}$$ $$x^3 cotx\over {1-cosx}$$

Solution : $$\displaystyle{\lim_{x \to 0}}$$ $$x^3 cosx\over {sinx(1-cosx)}$$ = $$\displaystyle{\lim_{x \to 0}}$$ $$x^3 cosx(1 + cosx)\over {sinxsin^2x}$$ = $$\displaystyle{\lim_{x \to 0}}$$ $${x^3\over sin^3x}.cosx(1 + cosx)$$ = 2 Similar Questions Evaluate the limit : $$\displaystyle{\lim_{x \to \infty}}$$ $$x^2 + x + 1\over {3x^2 + 2x – 5}$$ Evaluate : $$\displaystyle{\lim_{x \to \infty}}$$ $$({7x^2+1\over 5x^2-1})^{x^5\over {1-x^3}}$$ Evaluate …

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

Solution : Here f(x) = $${7x^2+1\over 5x^2-1}$$ $$\phi$$(x) = $${x^5\over {1-x^3}}$$ = $$x^2x^3\over 1-x^3$$ = $$x^2\over {1\over x^3}-1$$ $$\therefore$$ $$\displaystyle{\lim_{x \to \infty}}$$ f(x) = $$7\over 5$$ &amp;  $$\displaystyle{\lim_{x \to \infty}}$$ $$\phi$$(x) $$\rightarrow$$ – $$\infty$$ $$\implies$$ $$\displaystyle{\lim_{x \to \infty}}$$ $$(f(x))^{\phi (x)}$$ = $$({7\over 5})^{-\infty}$$ = 0 Similar Questions Evaluate : $$\displaystyle{\lim_{x \to 0}}$$ $$x^3 cotx\over {1-cosx}$$ …

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

Solution : $$\displaystyle{\lim_{x \to 0}}$$ $$2(sin(2+x)-sin2)+xsin(2+x)\over x$$ = $$\displaystyle{\lim_{x \to 0}}$$($$2.2.cos(2+{x\over 2})sin{x\over 2}\over x$$ + sin(2+x)) = $$\displaystyle{\lim_{x \to 0}}$$$$2cos(2+{x\over 2})sin{x\over 2}\over {x\over 2}$$ + $$\displaystyle{\lim_{x \to 0}}$$sin(2+x) = 2cos2 + sin2 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 : $$\displaystyle{\lim_{x \to 0}}$$ …

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

Solution : $$\displaystyle{\lim_{x \to \infty}}$$($${x^3+1\over x^2+1}-(ax+b)$$) = 2 $$\implies$$ $$\displaystyle{\lim_{x \to \infty}}$$$$x^3(1-a)-bx^2-ax+(1-b)\over x^2+1$$ = 2 $$\implies$$ $$\displaystyle{\lim_{x \to \infty}}$$$$x(1-a)-b-{a\over x}+{(1-b)\over x^2}\over 1+{1\over x^2}$$ = 2 $$\implies$$ 1 – a = 0, -b = 2 $$\implies$$ a = 1, b = -2 Similar Questions Evaluate : $$\displaystyle{\lim_{x \to 0}}$$ $$x^3 cotx\over {1-cosx}$$ Evaluate : \(\displaystyle{\lim_{x \to …