Question-33
Which of the following are true?
First Limit
\[ \begin{equation*} \lim\limits _{x\rightarrow \infty } \ \frac{x^{2}}{e^{x}} =0 \end{equation*} \]
\(\displaystyle e^{x}\) grows faster than \(\displaystyle x^{2}\). More concretely, we can use the Taylor series expansion for \(\displaystyle e^{x}\) and truncate it after the first four terms:
\[ \begin{equation*} e^{x} >1+x+\frac{x^{2}}{2} +\frac{x^{3}}{6} \Longrightarrow \frac{1}{e^{x}} < \frac{1}{1+x+\frac{x^{2}}{2} +\frac{x^{3}}{6}} \end{equation*} \]
Multiplying both sides by \(\displaystyle x^{2}\) and simplifying, we have the following inequality:
\[ \begin{equation*} 0< \frac{x^{2}}{e^{x}} < \frac{1}{\frac{1}{x^{2}} +\frac{1}{x} +\frac{1}{2} +\frac{x}{6}} \end{equation*} \]
As \(\displaystyle x\rightarrow \infty\), both the LHS and RHS tend to zero. We get the required limit by sandwich theorem.
Second Limit
\[ \begin{equation*} \begin{aligned} \lim\limits _{x\rightarrow 3} \ \frac{x^{2} -9}{x^{3} -27} & =\lim\limits _{x\rightarrow 3} \ \frac{( x-3)( x+3)}{( x-3)\left( x^{2} +3x+9\right)}\\ & \\ & =\lim\limits _{x\rightarrow 3} \ \frac{x+3}{x^{2} +3x+9}\\ & \\ & =\frac{2}{9} \end{aligned} \end{equation*} \]
Third Limit
\[ \begin{equation*} \lim\limits _{x\rightarrow -\infty } \ \frac{7}{e^{x}} =\infty \end{equation*} \]
As $x-$, the denominator tends to \(\displaystyle 0\) and the fraction tends to infinity.
Fourth Limit
\[ \begin{equation*} \begin{aligned} \lim\limits _{x\rightarrow -2} \ \ \frac{( x+2)^{2}}{4-x^{2}} & =\lim\limits _{x\rightarrow -2} \ \frac{x+2}{2-x} =0 \end{aligned} \end{equation*} \]