$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$\lim _ { h \rightarrow 0^ { + } } \frac { \sqrt { h ^ { 2 } + 13 h + 14 } - \sqrt { 14 } } { h }$
A) $\frac { 13 } { \sqrt { 28 } }$
B) $\frac { 13 } { 2 \sqrt { 14 } }$
C) $\frac { 13 } { 28 }$
D) Does not exist

Question

Quizplus · Accepted Answer

We can use the formula
$$\lim_{h\rightarrow 0}\frac{\sqrt{h^2+13h+14}-\sqrt{14}}{h}=\lim_{h\rightarrow 0} \frac{(\sqrt{h^2+13h+14}-\sqrt{14})(\sqrt{h^2+13h+14}+\sqrt{14})}{h(\sqrt{h^2+13h+14}+\sqrt{14})}$$
$$=\lim_{h\rightarrow 0}\frac{h^2+13h}{h(\sqrt{h^2+13h+14}+\sqrt{14})}=\lim_{h\rightarrow 0}\frac{h+13}{\sqrt{h^2+13h+14}+\sqrt{14}}$$
Now, if we let $h$ approach 0 from the right, then $h+13$ is positive and $\sqrt{h^2+13h+14}+\sqrt{14}$ is also positive, so by direct substitution, the limit equals $\frac{13}{2\sqrt{14}}$. Therefore, the answer is B.

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$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$