What is the limit as y approaches infinity of ((y^2-e^{(6y)}))/((4y^2+e^{7y))} ?

The limit as y approaches infinity of ((y^2-e^{(6y)}))/((4y^2+e^{7y))} is 0

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limit as y approaches infinity of ((y^2-e^{(6y)}))/((4y^2+e^{7y))}

y \to \infty lim (\frac{( y ^{2} - e ^{(6 y)} )}{( 4 y ^{2} + e ^{7 y} )})

0

Solution steps

y \to \infty lim (\frac{y ^{2} - e ^{6 y}}{4 y ^{2} + e ^{7 y}})

Divide by

e^{7 y} : \frac{\frac{y ^{2}}{e ^{7 y}} - \frac{1}{e ^{y}}}{\frac{4 y ^{2}}{e ^{7 y}} + 1}

= y \to \infty lim (\frac{\frac{y ^{2}}{e ^{7 y}} - \frac{1}{e ^{y}}}{\frac{4 y ^{2}}{e ^{7 y}} + 1})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{lim _{y \to \infty} ( \frac{y ^{2}}{e ^{7 y}} - \frac{1}{e ^{y}} )}{lim _{y \to \infty} ( \frac{4 y ^{2}}{e ^{7 y}} + 1 )}

y \to \infty lim (\frac{y ^{2}}{e ^{7 y}} - \frac{1}{e ^{y}}) = 0

y \to \infty lim (\frac{4 y ^{2}}{e ^{7 y}} + 1) = 1

= \frac{0}{1}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

What is the limit as y approaches infinity of ((y^2-e^{(6y)}))/((4y^2+e^{7y))} ?
The limit as y approaches infinity of ((y^2-e^{(6y)}))/((4y^2+e^{7y))} is 0