Find the constants $m$ and $n$ such that $\lim_{x \to 0} \sqrt{mx+n}-2x = 1$.

Understand the Problem

The question asks to find the constants $m$ and $n$ such that $\lim_{x \to 0} \sqrt{mx+n}-2x = 1$. We must find the values of $m$ and $n$ that satisfy the condition of the limit.

Answer

$m = 4$ $n = 1$

Steps to Solve

Evaluate the limit as x approaches 0

Substitute $x = 0$ into the expression $\sqrt{mx+n} - 2x$:

$\lim_{x \to 0} \sqrt{mx+n} - 2x = \sqrt{m(0)+n} - 2(0) = \sqrt{n}$

Set the limit equal to 1

We are given that the limit equals 1. Therefore, we have: $\sqrt{n} = 1$

Solve for $n$

Square both sides of the equation: $(\sqrt{n})^2 = 1^2$ $n = 1$

Rewrite the original limit expression Using the found value of n, we now have: $$ \lim_{x \to 0} \sqrt{mx+1} - 2x = 1 $$
Isolate the square root and square both sides $$ \lim_{x \to 0} \sqrt{mx+1} = 1 + 2x $$ $$ \lim_{x \to 0} (\sqrt{mx+1})^2 = (1 + 2x)^2 $$ $$ \lim_{x \to 0} mx+1 = 1 + 4x + 4x^2 $$
Simplify the equation and divide by x $$ mx+1 = 1 + 4x + 4x^2 $$ $$ mx = 4x + 4x^2 $$ $$ m = 4 + 4x $$
Take the limit as x approaches 0 $$ m = \lim_{x \to 0} (4 + 4x) $$ $$ m = 4 + 4(0) $$ $$ m = 4 $$
State the values of m and n We found $m = 4$ and $n = 1$.

$m = 4$ $n = 1$

More Information

The solution involves evaluating a limit and solving for the unknown constants $m$ and $n$. By directly substituting the limit value into the expression, we find a value for $n$. We then rearrange the equation, square both sides, and solve for $m$ by comparing coefficients or taking the limit again.

Tips

A common mistake is not recognizing that you can directly substitute x = 0 at the start to solve for n. Another mistake might be incorrectly squaring $(1+2x)$.

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