Find the degree in the limit

Notice that $\frac{(1)^n + 2(1) - 3}{1 - 1} = \frac{0}{0}$ . So we can simplify the top and bottom expression by L'Hopital's Rule; we differentiate both the numerator and the denominator. $\begin{aligned} \lim_{x \to 1} \frac{x^n + 2x - 3}{x - 1} &= \lim_{x \to 1} \frac{nx^{n-1} + 2}{1} \\ &= 2 + \lim_{x \to 1} nx^{n-1} \\ &= 2 + n(1)^{n-1} \\ &= 2 + n \end{aligned}$ Thus, $n + 2 = 25$ , and $n = \boxed{23}$ .

Differentiating the numerator and denominator (L'Hôpital's Rule), $\lim_{x \to 1} \dfrac{x^n + 2x - 3}{x - 1} = \lim_{x \to 1}\dfrac{nx^{n-1} + 2}{1}$ Now, it is convenient to plug-in $1$ to the limit. We are left with the following: $n\cdot 1^{(n-1)} + 2 = 25 \Rightarrow n + 2 = 25 \Rightarrow n = \boxed{23}$

*using l hospital method * and putting x=1 we get nx1+2=25 so n=23

by L' Hospitals rule we differentiate both numerator and denominator to get nx^(n-1)+2 and put x=1 to get n = 23

Use L'Hopital Rule. Differentiate both the numerator and denominator w.r.t. x

Let start this problem using L'Hôpital's rule

L'Hôpital's rule states that

$\lim_{x \to c} f(x)$ = $\lim_{x \to c} g(x)$ = 0 or $\pm \infty$ and,

$\lim_{x \to c}$ $\frac{f'(x)}{g'(x)}$ exists and $g'(x)$ $\not=$ 0 for all x,

then $\lim_{x \to c}$ $\frac{f(x)}{g(x)}$ = $\lim_{x \to c}$ $\frac{f'(x)}{g'(x)}$

So let Differentiate Numerator and Denominator individually,

So we are left with $\lim_{x \to 1}$ $nx^{n-1} + 2 = 25$

Replacing $x = 1$ we get $n + 2 = 25 \Longrightarrow n = \boxed{23}$

With L'Hospital ---> become {n.x^(n-1) + 2 }/ 1 = 25 ----> n*1 + 2 = 25 ---> n = 25 - 2 = 23. Answer : 23

If we replace x with 1 for every term involving x, we get an expression which looks like this: $\frac{1^n + 2 \times 1 - 3}{1 - 1} = 25$

This leaves us with 0 in the denominator which doesn't work, so we must use L'Hopital's rule to answer this question. L'hopital's rule basically says that if we replace the expression in the numerator ( $x^n + 2x - 3$ ) with it's derivative function ( $nx^{n-1} + 2$ ), and the expression in the denominator ( $x-1$ ) with it's own derivative function ( $1$ ), we will render the same limit value as we would had we evaluated the original expressions. Therefore we end up with this:

$\frac{nx^{n-1} + 2}{1} = 25$

which - as there is 1 in the denominator - can be written as:

$nx^{n^-1} + 2 = 25$

From here we can subtract two from both sides:

$nx^{n-1} = 23$

Now we can solve n. We have been told that as x approaches 1 the limit tends to 23, so we can go straight ahead and see whether substituting 1 into x works. It does - 1 to the power of any natural number such as $n$ is always equal to 1, and therefore $n \times 1^{n-1} = 23$ will easily render a value for n. All we need to do now is divide 23 by one (rearrange to $\frac{23}{x^{n-1}} = n$ ) leaving us with the answer:

$n = \boxed{23}$