# Find the limit of h ( t ) = 1 - t 2 2 t + 4 Find lim x 3 10- 27 x 4 x 4 + 8

## Question:

Find the limit of {eq}h(t)=\frac{1-t^2}{2t+4} {/eq}

Find {eq}\lim_{x\rightarrow\infty}\sqrt[3]{\frac{10-27x^4}{x^4+8}} {/eq}

## Definition of Limit at Infinity:

Assume that {eq}\displaystyle A {/eq} be subset of real numbers and {eq}\displaystyle f : A\to \mathbb{R} {/eq} be a real-valued function of {eq}x {/eq}, where {eq}\displaystyle\mathbb{R} {/eq} is the set of all real numbers. We say {eq}\displaystyle f {/eq} tends to {eq}\displaystyle L {/eq} as {eq}\displaystyle x\to \infty {/eq} if for every preassigned positive {eq}\displaystyle \epsilon {/eq} then there exists a real number {eq}\displaystyle G \in A {/eq} such that {eq}\displaystyle \left| {f\left( x \right) - L} \right| < \epsilon \,\,\,\,\forall \,\,x > G \ {/eq} and it is denoted by {eq}\displaystyle \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = L {/eq}. If there exists no such finite {eq}\displaystyle L {/eq}, we say that the limit does not exists {eq}x {/eq} tends to infinity.

Here we have {eq}\displaystyle h\left( x \right) = \sqrt[3]{{\frac{{10 - 27{x^4}}}{{{x^4} + 8}}}}. {/eq}

So the limit can be evaluated as follows:

{eq}\displaystyle \begin{align} \mathop {\lim }\limits_{x \to \infty } \sqrt[3]{{\frac{{10 - 27{x^4}}}{{{x^4} + 8}}}} &= \mathop {\lim }\limits_{x \to \infty } \sqrt[3]{{\frac{{\frac{{10 - 27{x^4}}}{{{x^4}}}}}{{\frac{{{x^4} + 8}}{{{x^4}}}}}}} \hfill \\ & = \mathop {\lim }\limits_{x \to \infty } \sqrt[3]{{\frac{{\frac{{10}}{{{x^4}}} - 27}}{{1 + \frac{8}{{{x^4}}}}}}} \hfill \\ & = {\left( {\frac{{27}}{8}} \right)^{\frac{1}{3}}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{as }}x \to \infty \Rightarrow x < {x^4} \to \infty \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^4}}} = 0} \right) \hfill \\ & = {\left\{ {{{\left( {\frac{3}{2}} \right)}^3}} \right\}^{\frac{1}{3}}} \hfill \\ & = \frac{3}{2}. \hfill \\ \end{align} {/eq}

Hence the value of the limit {eq}\displaystyle \mathop {\lim }\limits_{x \to \infty } \sqrt[3]{{\frac{{10 - 27{x^4}}}{{{x^4} + 8}}}} {/eq} is {eq}\displaystyle \boxed{\frac{3}{2}}. {/eq}