Limit as x goes to infinity of this problem?

Question

Find the limit as $x$ goes to $\infty$ of $\cfrac{\sqrt{x^2 + 16} + x}{x+3}$

I multiplied by the conjugate, but now I'm stuck. Now the denominator has a radical.

Answer 1

Setting $\dfrac1x=h,x^2+16=\dfrac{1+16h^2}{h^2}\implies\sqrt{x^2+16}=\dfrac{\sqrt{1+16h^2}}{|h|}$

As $h>0,|h|=+h$

$$\lim_{x\to\infty}\frac{\sqrt{x^2+16}+x}{x+3}=\lim_{h\to0^+}\frac{(\sqrt{1+16h^2}+1)h}{h(1+3h)}$$

Cancel out $h$ as $h\ne0$ as $h\to0$

Answer 2

In this case you just need to note that the highest power of $x$ occurring in numerator or denominator is $x^1$. Dividing top and bottom by this, $$\frac{\sqrt{x^2+16}+x}{x+3}=\frac{\sqrt{1+\frac{16}{x^2}}+1}{1+\frac3x}$$ and you should now be able to find the limit as $x\to\infty$.