NIMCET 2021 — Mathematics PYQ

Evaluation of the Limit 
The limit ${\lim }_{x\to \infty }{\left(\frac{x+7}{x+2}\right)}^{x+5}$ is evaluated using the standard limit form ${\lim }_{y\to \infty }{\left(1+\frac{k}{y}\right)}^y=e^k$. 
Step 1: Rewrite the Base of the Expression 
The base of the expression is rewritten to match the form $1+\frac{k}{y}$. $\frac{x+7}{x+2}=\frac{x+2+5}{x+2}=1+\frac{5}{x+2}$ 
Step 2: Adjust the Exponent 
The exponent is adjusted to match the denominator of the fraction in the base. Let $y=x+2$. As $x\to \infty$, it follows that $y\to \infty$. The exponent $x+5$ can be written in terms of $y$ as: $x+5=(y-2)+5=y+3$ 
Step 3: Substitute and Apply the Limit Property 
The expression is substituted with the new forms of the base and exponent. ${\lim }_{x\to \infty }{\left(1+\frac{5}{x+2}\right)}^{x+5}={\lim }_{y\to \infty }{\left(1+\frac{5}{y}\right)}^{y+3}$ This limit can be separated using exponent rules: ${\lim }_{y\to \infty }{\left(1+\frac{5}{y}\right)}^y\cdot {\left(1+\frac{5}{y}\right)}^3$ The limit of a product is the product of the limits: $\left({\lim }_{y\to \infty }{\left(1+\frac{5}{y}\right)}^y\right)\cdot \left({\lim }_{y\to \infty }{\left(1+\frac{5}{y}\right)}^3\right)$ The first part of the product is a standard limit form, which evaluates to $e^5$. The second part of the product evaluates to $1^3=1$ as ${\lim }_{y\to \infty }\frac{5}{y}=0$. $e^5\cdot 1=e^5$ 
Final Answer 
The final answer is \boxed{\text{e^5}}.