Question

Find $\mathop {\lim }\limits_{x \to \infty } \dfrac{{{{(2 + x)}^{40}}{{(4 + x)}^5}}}{{{{(2 - x)}^{45}}}}$
A. $- 1$
B. $1$
C. $16$
D. $32$

Answer 1

This is a question for solving of limit in which limit of $x$ is tending to infinity. To solve the limits tending to infinity, we divide both numerator and denominator with the highest power of the variable ($x$ in this question) and then substitute the limit to get the required answer.

Formula used:
Area of the trapezium =$\dfrac{1}{2}\left( a+b \right)\times h$, where ‘a’ and ‘b’ are the two parallel sides and ‘h’ is the height is of the trapezium.

Complete step by step solution:
In order to find the limit of the given function when limit of $x$ is tending to infinity, we will first find the highest power of the variable available in the given function.
In $\dfrac{{{{(2 + x)}^{40}}{{(4 + x)}^5}}}{{{{(2 - x)}^{45}}}}$, we can see that in the denominator the highest power of $x$ can be $45$ and in numerator there are two terms, in the first term the highest power of $x$ can be $40$ and in the second term the highest power of $x$ can be $5$, since both terms are in multiplication, so using law of indices for multiplication, we will get the highest power of $x$ equals $45$ in the numerator.
Now, the limit
$\mathop { = \lim }\limits_{x \to \infty } \dfrac{{{{(2 + x)}^{40}}{{(4 + x)}^5}}}{{{{(2 - x)}^{45}}}}$
So dividing both numerator and denominator with ${x^{45}}$, we will get
$= \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{{{{(2 + x)}^{40}}{{(4 + x)}^5}}}{{{x^{45}}}}}}{{\dfrac{{{{(2 - x)}^{45}}}}{{{x^{45}}}}}}$
Using law of indices for multiplication, we can write it as
$= \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{{{{(2 + x)}^{40}}{{(4 + x)}^5}}}{{{x^{40}} \times {x^5}}}}}{{\dfrac{{{{(2 - x)}^{45}}}}{{{x^{45}}}}}}$
Now, combining terms having same exponents, we will get
$= \mathop {\lim }\limits_{x \to \infty } \dfrac{{{{\left( {\dfrac{{2 + x}}{x}} \right)}^{40}}{{\left( {\dfrac{{4 + x}}{x}} \right)}^5}}}{{{{\left( {\dfrac{{2 - x}}{x}} \right)}^{45}}}} \\\ = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{{\left( {\dfrac{2}{x} + 1} \right)}^{40}}{{\left( {\dfrac{4}{x} + 1} \right)}^5}}}{{{{\left( {\dfrac{2}{x} - 1} \right)}^{45}}}} \\\$
Now using property of limit, we can write
$= \dfrac{{\mathop {\lim }\limits_{x \to \infty } {{\left( {\dfrac{2}{x} + 1} \right)}^{40}}\mathop {\lim }\limits_{x \to \infty } {{\left( {\dfrac{4}{x} + 1} \right)}^5}}}{{\mathop {\lim }\limits_{x \to \infty } {{\left( {\dfrac{2}{x} - 1} \right)}^{45}}}}$
Solving the limits, we will get

$$= \dfrac{{{{\left( {0 + 1} \right)}^{40}}{{\left( {0 + 1} \right)}^5}}}{{{{\left( {0 - 1} \right)}^{45}}}} \\\ = \dfrac{{1 \times 1}}{{ - 1}} \\\ = - 1 \\\$$

So, the correct answer is Option A.

Note: You can solve or answer this type of limit related questions directly, by seeing the highest powers of numerator and denominator and also their coefficients. If the highest power of denominator is greater than that of numerator then limit will be zero, and if equal then the limit is ratio of coefficient of highest power of numerator and denominator.