question evaluate the following limit using lhospitals rule. lim x→0 7 sin(5x) / 2 tan(2x) enter an exact…

Explanation:

Step1: Check the form of the limit

As (x\rightarrow0), (\sin(5x)\rightarrow0) and (\tan(2x)\rightarrow0), so it's in (\frac{0}{0}) form and L'Hopital's rule can be applied.

Step2: Differentiate the numerator and denominator

The derivative of (y = 7\sin(5x)) using the chain - rule ((u = 5x,y = 7\sin(u))) is (y^\prime=7\times5\cos(5x)=35\cos(5x)). The derivative of (y = 2\tan(2x)) using the chain - rule ((u = 2x,y = 2\tan(u))) is (y^\prime=2\times2\sec^{2}(2x)=4\sec^{2}(2x)).

Step3: Evaluate the new limit

We now find (\lim_{x\rightarrow0}\frac{35\cos(5x)}{4\sec^{2}(2x)}). Substitute (x = 0) into the expression. Since (\cos(0)=1) and (\sec(0)=1), we have (\frac{35\times\cos(0)}{4\times\sec^{2}(0)}=\frac{35\times1}{4\times1^{2}}=\frac{35}{4}).

Answer:

(\frac{35}{4})