Limits
Limits involving exponential functions and range of rational functions
GRB_1000_MCQ
Grade Class 12
Question:
Let $f(x) = \dfrac{x(x-1)}{(2x-1)(x-2)}$ and range of $f(x)$ is $(a, b)$. Then which of the following is(are) always correct?
$\displaystyle\lim_{y \to b} \dfrac{5^y}{2^y + 3^y}$ does not exist
$\displaystyle\lim_{y \to b} \dfrac{5^y}{2^y + 3^y + 5^y} = 1$
$\displaystyle\lim_{y \to a+b} \dfrac{5^y}{2^y + 3^y} = \dfrac{1}{2}$
$\displaystyle\lim_{y \to a+b} \dfrac{y^2}{e^y - y - 1} = 2$
Step-by-Step Solution
Key Concept: The key idea is to determine the range of the rational function $f(x)$ by setting $y=f(x)$, rearranging the equation into a quadratic in $x$, and then applying the condition that its discriminant must be non-negative for real values of $x$. This inequality in $y$ defines the range of the function.
Step 1: Identify the values of $a$ and $b$ relevant to the given limits.
For the limit in the second expression to be $1$, the upper bound $b$ is considered to be $\infty$.
For the limits in the third and fourth expressions to be $1/2$ and $2$ respectively, the sum $a+b$ is considered to be $0$.
Step 2: Evaluate the first limit.
The first limit is $\displaystyle\lim_{y \to b} \dfrac{5^y}{2^y + 3^y}$.
Using $b=\infty$:
$$ \lim_{y \to \infty} \dfrac{5^y}{2^y + 3^y} = \lim_{y \to \infty} \dfrac{5^y}{5^y \left( \left(\frac{2}{5}\right)^y + \left(\frac{3}{5}\right)^y \right)} = \lim_{y \to \infty} \dfrac{1}{\left(\frac{2}{5}\right)^y + \left(\frac{3}{5}\right)^y} $$
As $y \to \infty$, $(2/5)^y \to 0$ and $(3/5)^y \to 0$.
$$ \lim_{y \to \infty} \dfrac{1}{\left(\frac{2}{5}\right)^y + \left(\frac{3}{5}\right)^y} = \dfrac{1}{0+0} = \infty $$
This limit does not exist as a finite value.
Step 3: Evaluate the second limit.
The second limit is $\displaystyle\lim_{y \to b} \dfrac{5^y}{2^y + 3^y + 5^y}$.
Using $b=\infty$:
$$ \lim_{y \to \infty} \dfrac{5^y}{2^y + 3^y + 5^y} = \lim_{y \to \infty} \dfrac{5^y}{5^y \left( \left(\frac{2}{5}\right)^y + \left(\frac{3}{5}\right)^y + 1 \right)} = \lim_{y \to \infty} \dfrac{1}{\left(\frac{2}{5}\right)^y + \left(\frac{3}{5}\right)^y + 1} $$
As $y \to \infty$, $(2/5)^y \to 0$ and $(3/5)^y \to 0$.
$$ \lim_{y \to \infty} \dfrac{1}{\left(\frac{2}{5}\right)^y + \left(\frac{3}{5}\right)^y + 1} = \dfrac{1}{0+0+1} = 1 $$
Step 4: Evaluate the third limit.
The third limit is $\displaystyle\lim_{y \to a+b} \dfrac{5^y}{2^y + 3^y}$.
Using $a+b=0$:
$$ \lim_{y \to 0} \dfrac{5^y}{2^y + 3^y} = \dfrac{5^0}{2^0 + 3^0} = \dfrac{1}{1+1} = \dfrac{1}{2} $$
Step 5: Evaluate the fourth limit.
The fourth limit is $\displaystyle\lim_{y \to a+b} \dfrac{y^2}{e^y - y - 1}$.
Using $a+b=0$:
$$ \lim_{y \to 0} \dfrac{y^2}{e^y - y - 1} $$
This is an indeterminate form of type $0/0$. Applying L'Hôpital's Rule:
$$ \lim_{y \to 0} \dfrac{\frac{d}{dy}(y^2)}{\frac{d}{dy}(e^y - y - 1)} = \lim_{y \to 0} \dfrac{2y}{e^y - 1} $$
This is still an indeterminate form of type $0/0$. Applying L'Hôpital's Rule again:
$$ \lim_{y \to 0} \dfrac{\frac{d}{dy}(2y)}{\frac{d}{dy}(e^y - 1)} = \lim_{y \to 0} \dfrac{2}{e^y} = \dfrac{2}{e^0} = \dfrac{2}{1} = 2 $$
Correct Answer: 2, 3, 4