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Question: If \[{\text{x > 0}}\] and \[{\text{g}}\] is a bounded function, then \[\mathop {\lim }\limits_{x \to...

If x > 0{\text{x > 0}} and g{\text{g}} is a bounded function, then lim⁡x→∞f(y) enx + g(y)enx + 1\mathop {\lim }\limits_{x \to \infty } \dfrac{{{\text{f(y) }}{{\text{e}}^{{\text{nx}}}}{\text{ + g(y)}}}}{{{{\text{e}}^{{\text{nx}}}}{\text{ + 1}}}} is where y{\text{y}} is independent of x{\text{x}}.
A. 0
B. g(y){\text{g(y)}}
C. f(y){\text{f(y)}}
D.  - g(y){\text{ - g(y)}}

Explanation

Solution

In this question, we need to choose the correct option from the given options. By solving the given term to find the required solution. From that we will get the correct option. To solve the given we have to apply the limit which is in the given. And more we want to simplify them. After Simplification we will get the result which is required.

Complete step-by-step answer:
We have given in the question that,
lim⁡x→∞f(y) enx + g(y)enx + 1\mathop {\lim }\limits_{x \to \infty } \dfrac{{{\text{f(y) }}{{\text{e}}^{{\text{nx}}}}{\text{ + g(y)}}}}{{{{\text{e}}^{{\text{nx}}}}{\text{ + 1}}}}
So, we are going to solve the problem using this
Now we will expand and write the denominator part,
After expanding we will get that,
 = lim⁡x→∞f(y) enxenx + 1 + g(y)enx + 1{\text{ = }}\mathop {\lim }\limits_{x \to \infty } \dfrac{{{\text{f(y) }}{{\text{e}}^{{\text{nx}}}}}}{{{{\text{e}}^{{\text{nx}}}}{\text{ + 1}}}}{\text{ + }}\dfrac{{{\text{g(y)}}}}{{{{\text{e}}^{{\text{nx}}}}{\text{ + 1}}}}
By Bringing enx{{\text{e}}^{{\text{nx}}}} to the denominator part we will have
 = lim⁡x→∞f(y) enx + 1enx + g(y)enx + 1{\text{ = }}\mathop {\lim }\limits_{x \to \infty } \dfrac{{{\text{f(y) }}}}{{\dfrac{{{{\text{e}}^{{\text{nx}}}}{\text{ + 1}}}}{{{{\text{e}}^{{\text{nx}}}}}}}}{\text{ + }}\dfrac{{{\text{g(y)}}}}{{{{\text{e}}^{{\text{nx}}}}{\text{ + 1}}}}
Here we are rewriting the above, so we will get that
{\text{ = }}$$$${\text{ }}\mathop {\lim }\limits_{x \to \infty } \dfrac{{{\text{f(y) }}}}{{{\text{1 + }}\dfrac{1}{{{{\text{e}}^{{\text{nx}}}}}}}}{\text{ + }}\dfrac{{{\text{g(y)}}}}{{{{\text{e}}^{{\text{nx}}}}{\text{ + 1}}}}
Here, we are applying the limit which is given,
After applying the limit, we will get that
 = f(y) 1+1e∞ + g(y)e∞ + 1{\text{ = }}\dfrac{{{\text{f(y) }}}}{{1 + \dfrac{1}{{{{\text{e}}^\infty }}}}}{\text{ + }}\dfrac{{{\text{g(y)}}}}{{{{\text{e}}^\infty }{\text{ + 1}}}}
Here, there is an important thing we are going to substitute the value
Since, we already know that \dfrac{{\text{1}}}{{{{\text{e}}^\infty }}}$$$$ = 0,
Also, we know that e∞=∞{{\text{e}}^\infty } = \infty
By applying these things that we know, on the above, we will have that
{\text{ = }}$$$$\dfrac{{{\text{f(y)}}}}{{{\text{1 + 0}}}}{\text{ + }}\dfrac{{{\text{g(y)}}}}{\infty }
We already know that any number dividing by 1{\text{1}} is the same number
Also, we know anything divided by ∞\infty is zero
So, we will get that
{\text{ = }}$$$${\text{f(y) + 0}}
When you add this, you will get
{\text{ = }}$$$${\text{f(y)}}
Therefore, we had got the required result that lim⁡x→∞f(y) enx + g(y)enx + 1\mathop {\lim }\limits_{x \to \infty } \dfrac{{{\text{f(y) }}{{\text{e}}^{{\text{nx}}}}{\text{ + g(y)}}}}{{{{\text{e}}^{{\text{nx}}}}{\text{ + 1}}}} is equal to f(y){\text{f(y)}}.

Option C is the correct answer.

Note: A function defined on some set with real or complex values is called bounded if the set of its values is bounded.
The correct way to learn is to understand the concept thoroughly.
And applying limit value \dfrac{{\text{1}}}{{{{\text{e}}^\infty }}}$$$$ = 0, e∞=∞{{\text{e}}^\infty } = \infty correctly. You must know the limit value properly.
Also, we need to be careful, on the ways that, how to split and rewrite them correctly and applying the values according to that.