Limits of functions as x approaches a

Question

I need help to find the limits of these two functions : $$\lim_{x\to a}\frac{x^n-a^n}{x^p-a^p}$$ where $n,p$ are integers.

And : $$\lim_{x\to a}{\frac{x\sin(a)-a\sin(x)}{x-a\log_a(x)}}$$ where $a>0$ and $a\notin{{1,e}}$

I can't use L'Hospital rule, any help would be very appreciated !

Answer 1

The first one simply obtain by L"Hospital rule: $$\lim_{x\to a}\dfrac{x^n-a^n}{x^p-a^p}=\lim_{x\to a}\dfrac{nx^{n-1}}{px^{p-1}}=\dfrac{n}{p}a^{n-p}$$

Edit: By identity $$\lim_{x\to a}\dfrac{x^n-a^n}{x^p-a^p}=\lim_{x\to a}\dfrac{(x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots+a^{n-2}x+a^{n-1}}{(x-a)(x^{p-1}+ax^{p-2}+a^2x^{p-3}+\cdots+a^{p-2}x+a^{p-1}}=\dfrac{n}{p}a^{n-p}$$

Answer 2

$$\frac{x^n-a^n}{x^p-a^p}=\frac{x^n-a^n}{x-a}\cdot\frac{x-a}{x^p-a^p}\xrightarrow[x\to a]{}(x^n)'|_{x=a}\cdot\frac1{(x^p)'|_{x=a}}=\frac{na^{n-1}}{pa^{p-1}}=\frac npa^{n-p}$$

Answer 3

HINT

By $f(x)=x^n$ and $g(x)=x^p$ we have

$$\lim_{x\to a}\frac{x^n-a^n}{x^p-a^p}=\lim_{x\to a}\frac{x^n-a^n}{x-a}\frac{x-a}{x^p-a^p}=\frac{f'(a)}{g'(a)}$$

and by $f(x)=x\sin(a)-a\sin(x)$ and $g(x)=a-a\log_a(x)$ we have

$$\lim_{x\to a}{\frac{x\sin(a)-a\sin(x)}{a-a\log_a(x)}}=\lim_{x\to a}{\frac{x\sin(a)-a\sin(x)}{x-a}}{\frac{x-a}{a-a\log_a(x)}}=\frac{f'(a)}{g'(a)}$$

Answer 4

For the first one you can factor top and bottom and cancel $(x-a)$

For the second one the answer is straight forward because the top goes to zero and the bottom does not.