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

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

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

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.

Leave a Comment

Your email address will not be published.